• PRO Courses Guides New Tech Help Pro Expert Videos About wikiHow Pro Upgrade Sign In
  • EDIT Edit this Article
  • EXPLORE Tech Help Pro About Us Random Article Quizzes Request a New Article Community Dashboard This Or That Game Popular Categories Arts and Entertainment Artwork Books Movies Computers and Electronics Computers Phone Skills Technology Hacks Health Men's Health Mental Health Women's Health Relationships Dating Love Relationship Issues Hobbies and Crafts Crafts Drawing Games Education & Communication Communication Skills Personal Development Studying Personal Care and Style Fashion Hair Care Personal Hygiene Youth Personal Care School Stuff Dating All Categories Arts and Entertainment Finance and Business Home and Garden Relationship Quizzes Cars & Other Vehicles Food and Entertaining Personal Care and Style Sports and Fitness Computers and Electronics Health Pets and Animals Travel Education & Communication Hobbies and Crafts Philosophy and Religion Work World Family Life Holidays and Traditions Relationships Youth
  • Browse Articles
  • Learn Something New
  • Quizzes Hot
  • This Or That Game New
  • Train Your Brain
  • Explore More
  • Support wikiHow
  • About wikiHow
  • Log in / Sign up
  • Education and Communications
  • Classical Mechanics

How to Solve a Projectile Motion Problem

Last Updated: October 6, 2017

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, volunteer authors worked to edit and improve it over time. This article has been viewed 73,306 times. Learn more...

Projectile motion is often one of the most difficult topics to understand in physics classes. Most of the time, there is not a direct way to get the answer; you need to solve for a few other variables to get the answer you are looking for. This means in order to find the distance an object traveled, you might first have to find the time it took or the initial velocity first. Just follow these steps and you should be able to fly through projectile motion problems!

Step 1 Determine what type of problem it is.

  • (1) an object is thrown off a higher ground than what it will land on.
  • (2) the object starts on the ground, soars through the air, and then lands on the ground some distance away from where it started.

Step 2 Draw a picture.

Community Q&A

Community Answer

You Might Also Like

Calculate Acceleration

About This Article

  • Send fan mail to authors

Did this article help you?

Am I a Narcissist or an Empath Quiz

Featured Articles

Ask Better Questions

Trending Articles

Everything You Need to Know to Rock the Corporate Goth Aesthetic

Watch Articles

Cook Fresh Cauliflower

  • Terms of Use
  • Privacy Policy
  • Do Not Sell or Share My Info
  • Not Selling Info

Get all the best how-tos!

Sign up for wikiHow's weekly email newsletter

Youtube

  • TPC and eLearning
  • What's NEW at TPC?
  • Read Watch Interact
  • Practice Review Test
  • Teacher-Tools
  • Subscription Selection
  • Seat Calculator
  • Ad Free Account
  • Edit Profile Settings
  • Classes (Version 2)
  • Student Progress Edit
  • Task Properties
  • Export Student Progress
  • Task, Activities, and Scores
  • Metric Conversions Questions
  • Metric System Questions
  • Metric Estimation Questions
  • Significant Digits Questions
  • Proportional Reasoning
  • Acceleration
  • Distance-Displacement
  • Dots and Graphs
  • Graph That Motion
  • Match That Graph
  • Name That Motion
  • Motion Diagrams
  • Pos'n Time Graphs Numerical
  • Pos'n Time Graphs Conceptual
  • Up And Down - Questions
  • Balanced vs. Unbalanced Forces
  • Change of State
  • Force and Motion
  • Mass and Weight
  • Match That Free-Body Diagram
  • Net Force (and Acceleration) Ranking Tasks
  • Newton's Second Law
  • Normal Force Card Sort
  • Recognizing Forces
  • Air Resistance and Skydiving
  • Solve It! with Newton's Second Law
  • Which One Doesn't Belong?
  • Component Addition Questions
  • Head-to-Tail Vector Addition
  • Projectile Mathematics
  • Trajectory - Angle Launched Projectiles
  • Trajectory - Horizontally Launched Projectiles
  • Vector Addition
  • Vector Direction
  • Which One Doesn't Belong? Projectile Motion
  • Forces in 2-Dimensions
  • Being Impulsive About Momentum
  • Explosions - Law Breakers
  • Hit and Stick Collisions - Law Breakers
  • Case Studies: Impulse and Force
  • Impulse-Momentum Change Table
  • Keeping Track of Momentum - Hit and Stick
  • Keeping Track of Momentum - Hit and Bounce
  • What's Up (and Down) with KE and PE?
  • Energy Conservation Questions
  • Energy Dissipation Questions
  • Energy Ranking Tasks
  • LOL Charts (a.k.a., Energy Bar Charts)
  • Match That Bar Chart
  • Words and Charts Questions
  • Name That Energy
  • Stepping Up with PE and KE Questions
  • Case Studies - Circular Motion
  • Circular Logic
  • Forces and Free-Body Diagrams in Circular Motion
  • Gravitational Field Strength
  • Universal Gravitation
  • Angular Position and Displacement
  • Linear and Angular Velocity
  • Angular Acceleration
  • Rotational Inertia
  • Balanced vs. Unbalanced Torques
  • Getting a Handle on Torque
  • Torque-ing About Rotation
  • Properties of Matter
  • Fluid Pressure
  • Buoyant Force
  • Sinking, Floating, and Hanging
  • Pascal's Principle
  • Flow Velocity
  • Balloon Interactions
  • Charge and Charging
  • Charge Interactions
  • Charging by Induction
  • Conductors and Insulators
  • Coulombs Law
  • Electric Field
  • Electric Field Intensity
  • Polarization
  • Case Studies: Electric Power
  • Know Your Potential
  • Light Bulb Anatomy
  • I = ∆V/R Equations as a Guide to Thinking
  • Parallel Circuits - ∆V = I•R Calculations
  • Resistance Ranking Tasks
  • Series Circuits - ∆V = I•R Calculations
  • Series vs. Parallel Circuits
  • Equivalent Resistance
  • Period and Frequency of a Pendulum
  • Pendulum Motion: Velocity and Force
  • Energy of a Pendulum
  • Period and Frequency of a Mass on a Spring
  • Horizontal Springs: Velocity and Force
  • Vertical Springs: Velocity and Force
  • Energy of a Mass on a Spring
  • Decibel Scale
  • Frequency and Period
  • Closed-End Air Columns
  • Name That Harmonic: Strings
  • Rocking the Boat
  • Wave Basics
  • Matching Pairs: Wave Characteristics
  • Wave Interference
  • Waves - Case Studies
  • Color Addition and Subtraction
  • Color Filters
  • If This, Then That: Color Subtraction
  • Light Intensity
  • Color Pigments
  • Converging Lenses
  • Curved Mirror Images
  • Law of Reflection
  • Refraction and Lenses
  • Total Internal Reflection
  • Who Can See Who?
  • Formulas and Atom Counting
  • Atomic Models
  • Bond Polarity
  • Entropy Questions
  • Cell Voltage Questions
  • Heat of Formation Questions
  • Reduction Potential Questions
  • Oxidation States Questions
  • Measuring the Quantity of Heat
  • Hess's Law
  • Oxidation-Reduction Questions
  • Galvanic Cells Questions
  • Thermal Stoichiometry
  • Molecular Polarity
  • Quantum Mechanics
  • Balancing Chemical Equations
  • Bronsted-Lowry Model of Acids and Bases
  • Classification of Matter
  • Collision Model of Reaction Rates
  • Density Ranking Tasks
  • Dissociation Reactions
  • Complete Electron Configurations
  • Enthalpy Change Questions
  • Equilibrium Concept
  • Equilibrium Constant Expression
  • Equilibrium Calculations - Questions
  • Equilibrium ICE Table
  • Ionic Bonding
  • Lewis Electron Dot Structures
  • Line Spectra Questions
  • Measurement and Numbers
  • Metals, Nonmetals, and Metalloids
  • Metric Estimations
  • Metric System
  • Molarity Ranking Tasks
  • Mole Conversions
  • Name That Element
  • Names to Formulas
  • Names to Formulas 2
  • Nuclear Decay
  • Particles, Words, and Formulas
  • Periodic Trends
  • Precipitation Reactions and Net Ionic Equations
  • Pressure Concepts
  • Pressure-Temperature Gas Law
  • Pressure-Volume Gas Law
  • Chemical Reaction Types
  • Significant Digits and Measurement
  • States Of Matter Exercise
  • Stoichiometry - Math Relationships
  • Subatomic Particles
  • Spontaneity and Driving Forces
  • Gibbs Free Energy
  • Volume-Temperature Gas Law
  • Acid-Base Properties
  • Energy and Chemical Reactions
  • Chemical and Physical Properties
  • Valence Shell Electron Pair Repulsion Theory
  • Writing Balanced Chemical Equations
  • Mission CG1
  • Mission CG10
  • Mission CG2
  • Mission CG3
  • Mission CG4
  • Mission CG5
  • Mission CG6
  • Mission CG7
  • Mission CG8
  • Mission CG9
  • Mission EC1
  • Mission EC10
  • Mission EC11
  • Mission EC12
  • Mission EC2
  • Mission EC3
  • Mission EC4
  • Mission EC5
  • Mission EC6
  • Mission EC7
  • Mission EC8
  • Mission EC9
  • Mission RL1
  • Mission RL2
  • Mission RL3
  • Mission RL4
  • Mission RL5
  • Mission RL6
  • Mission KG7
  • Mission RL8
  • Mission KG9
  • Mission RL10
  • Mission RL11
  • Mission RM1
  • Mission RM2
  • Mission RM3
  • Mission RM4
  • Mission RM5
  • Mission RM6
  • Mission RM8
  • Mission RM10
  • Mission LC1
  • Mission RM11
  • Mission LC2
  • Mission LC3
  • Mission LC4
  • Mission LC5
  • Mission LC6
  • Mission LC8
  • Mission SM1
  • Mission SM2
  • Mission SM3
  • Mission SM4
  • Mission SM5
  • Mission SM6
  • Mission SM8
  • Mission SM10
  • Mission KG10
  • Mission SM11
  • Mission KG2
  • Mission KG3
  • Mission KG4
  • Mission KG5
  • Mission KG6
  • Mission KG8
  • Mission KG11
  • Mission F2D1
  • Mission F2D2
  • Mission F2D3
  • Mission F2D4
  • Mission F2D5
  • Mission F2D6
  • Mission KC1
  • Mission KC2
  • Mission KC3
  • Mission KC4
  • Mission KC5
  • Mission KC6
  • Mission KC7
  • Mission KC8
  • Mission AAA
  • Mission SM9
  • Mission LC7
  • Mission LC9
  • Mission NL1
  • Mission NL2
  • Mission NL3
  • Mission NL4
  • Mission NL5
  • Mission NL6
  • Mission NL7
  • Mission NL8
  • Mission NL9
  • Mission NL10
  • Mission NL11
  • Mission NL12
  • Mission MC1
  • Mission MC10
  • Mission MC2
  • Mission MC3
  • Mission MC4
  • Mission MC5
  • Mission MC6
  • Mission MC7
  • Mission MC8
  • Mission MC9
  • Mission RM7
  • Mission RM9
  • Mission RL7
  • Mission RL9
  • Mission SM7
  • Mission SE1
  • Mission SE10
  • Mission SE11
  • Mission SE12
  • Mission SE2
  • Mission SE3
  • Mission SE4
  • Mission SE5
  • Mission SE6
  • Mission SE7
  • Mission SE8
  • Mission SE9
  • Mission VP1
  • Mission VP10
  • Mission VP2
  • Mission VP3
  • Mission VP4
  • Mission VP5
  • Mission VP6
  • Mission VP7
  • Mission VP8
  • Mission VP9
  • Mission WM1
  • Mission WM2
  • Mission WM3
  • Mission WM4
  • Mission WM5
  • Mission WM6
  • Mission WM7
  • Mission WM8
  • Mission WE1
  • Mission WE10
  • Mission WE2
  • Mission WE3
  • Mission WE4
  • Mission WE5
  • Mission WE6
  • Mission WE7
  • Mission WE8
  • Mission WE9
  • Vector Walk Interactive
  • Name That Motion Interactive
  • Kinematic Graphing 1 Concept Checker
  • Kinematic Graphing 2 Concept Checker
  • Graph That Motion Interactive
  • Rocket Sled Concept Checker
  • Force Concept Checker
  • Free-Body Diagrams Concept Checker
  • Free-Body Diagrams The Sequel Concept Checker
  • Skydiving Concept Checker
  • Elevator Ride Concept Checker
  • Vector Addition Concept Checker
  • Vector Walk in Two Dimensions Interactive
  • Name That Vector Interactive
  • River Boat Simulator Concept Checker
  • Projectile Simulator 2 Concept Checker
  • Projectile Simulator 3 Concept Checker
  • Turd the Target 1 Interactive
  • Turd the Target 2 Interactive
  • Balance It Interactive
  • Go For The Gold Interactive
  • Egg Drop Concept Checker
  • Fish Catch Concept Checker
  • Exploding Carts Concept Checker
  • Collision Carts - Inelastic Collisions Concept Checker
  • Its All Uphill Concept Checker
  • Stopping Distance Concept Checker
  • Chart That Motion Interactive
  • Roller Coaster Model Concept Checker
  • Uniform Circular Motion Concept Checker
  • Horizontal Circle Simulation Concept Checker
  • Vertical Circle Simulation Concept Checker
  • Race Track Concept Checker
  • Gravitational Fields Concept Checker
  • Orbital Motion Concept Checker
  • Balance Beam Concept Checker
  • Torque Balancer Concept Checker
  • Aluminum Can Polarization Concept Checker
  • Charging Concept Checker
  • Name That Charge Simulation
  • Coulomb's Law Concept Checker
  • Electric Field Lines Concept Checker
  • Put the Charge in the Goal Concept Checker
  • Circuit Builder Concept Checker (Series Circuits)
  • Circuit Builder Concept Checker (Parallel Circuits)
  • Circuit Builder Concept Checker (∆V-I-R)
  • Circuit Builder Concept Checker (Voltage Drop)
  • Equivalent Resistance Interactive
  • Pendulum Motion Simulation Concept Checker
  • Mass on a Spring Simulation Concept Checker
  • Particle Wave Simulation Concept Checker
  • Boundary Behavior Simulation Concept Checker
  • Slinky Wave Simulator Concept Checker
  • Simple Wave Simulator Concept Checker
  • Wave Addition Simulation Concept Checker
  • Standing Wave Maker Simulation Concept Checker
  • Color Addition Concept Checker
  • Painting With CMY Concept Checker
  • Stage Lighting Concept Checker
  • Filtering Away Concept Checker
  • InterferencePatterns Concept Checker
  • Young's Experiment Interactive
  • Plane Mirror Images Interactive
  • Who Can See Who Concept Checker
  • Optics Bench (Mirrors) Concept Checker
  • Name That Image (Mirrors) Interactive
  • Refraction Concept Checker
  • Total Internal Reflection Concept Checker
  • Optics Bench (Lenses) Concept Checker
  • Kinematics Preview
  • Velocity Time Graphs Preview
  • Moving Cart on an Inclined Plane Preview
  • Stopping Distance Preview
  • Cart, Bricks, and Bands Preview
  • Fan Cart Study Preview
  • Friction Preview
  • Coffee Filter Lab Preview
  • Friction, Speed, and Stopping Distance Preview
  • Up and Down Preview
  • Projectile Range Preview
  • Ballistics Preview
  • Juggling Preview
  • Marshmallow Launcher Preview
  • Air Bag Safety Preview
  • Colliding Carts Preview
  • Collisions Preview
  • Engineering Safer Helmets Preview
  • Push the Plow Preview
  • Its All Uphill Preview
  • Energy on an Incline Preview
  • Modeling Roller Coasters Preview
  • Hot Wheels Stopping Distance Preview
  • Ball Bat Collision Preview
  • Energy in Fields Preview
  • Weightlessness Training Preview
  • Roller Coaster Loops Preview
  • Universal Gravitation Preview
  • Keplers Laws Preview
  • Kepler's Third Law Preview
  • Charge Interactions Preview
  • Sticky Tape Experiments Preview
  • Wire Gauge Preview
  • Voltage, Current, and Resistance Preview
  • Light Bulb Resistance Preview
  • Series and Parallel Circuits Preview
  • Thermal Equilibrium Preview
  • Linear Expansion Preview
  • Heating Curves Preview
  • Electricity and Magnetism - Part 1 Preview
  • Electricity and Magnetism - Part 2 Preview
  • Vibrating Mass on a Spring Preview
  • Period of a Pendulum Preview
  • Wave Speed Preview
  • Slinky-Experiments Preview
  • Standing Waves in a Rope Preview
  • Sound as a Pressure Wave Preview
  • DeciBel Scale Preview
  • DeciBels, Phons, and Sones Preview
  • Sound of Music Preview
  • Shedding Light on Light Bulbs Preview
  • Models of Light Preview
  • Electromagnetic Radiation Preview
  • Electromagnetic Spectrum Preview
  • EM Wave Communication Preview
  • Digitized Data Preview
  • Light Intensity Preview
  • Concave Mirrors Preview
  • Object Image Relations Preview
  • Snells Law Preview
  • Reflection vs. Transmission Preview
  • Magnification Lab Preview
  • Reactivity Preview
  • Ions and the Periodic Table Preview
  • Periodic Trends Preview
  • Gaining Teacher Access
  • Tasks and Classes
  • Tasks - Classic
  • Subscription
  • Subscription Locator
  • 1-D Kinematics
  • Newton's Laws
  • Vectors - Motion and Forces in Two Dimensions
  • Momentum and Its Conservation
  • Work and Energy
  • Circular Motion and Satellite Motion
  • Thermal Physics
  • Static Electricity
  • Electric Circuits
  • Vibrations and Waves
  • Sound Waves and Music
  • Light and Color
  • Reflection and Mirrors
  • About the Physics Interactives
  • Task Tracker
  • Usage Policy
  • Newtons Laws
  • Vectors and Projectiles
  • Forces in 2D
  • Momentum and Collisions
  • Circular and Satellite Motion
  • Balance and Rotation
  • Electromagnetism
  • Waves and Sound
  • Forces in Two Dimensions
  • Work, Energy, and Power
  • Circular Motion and Gravitation
  • Sound Waves
  • 1-Dimensional Kinematics
  • Circular, Satellite, and Rotational Motion
  • Einstein's Theory of Special Relativity
  • Waves, Sound and Light
  • QuickTime Movies
  • About the Concept Builders
  • Pricing For Schools
  • Directions for Version 2
  • Measurement and Units
  • Relationships and Graphs
  • Rotation and Balance
  • Vibrational Motion
  • Reflection and Refraction
  • Teacher Accounts
  • Task Tracker Directions
  • Kinematic Concepts
  • Kinematic Graphing
  • Wave Motion
  • Sound and Music
  • About CalcPad
  • 1D Kinematics
  • Vectors and Forces in 2D
  • Simple Harmonic Motion
  • Rotational Kinematics
  • Rotation and Torque
  • Rotational Dynamics
  • Electric Fields, Potential, and Capacitance
  • Transient RC Circuits
  • Light Waves
  • Units and Measurement
  • Stoichiometry
  • Molarity and Solutions
  • Thermal Chemistry
  • Acids and Bases
  • Kinetics and Equilibrium
  • Solution Equilibria
  • Oxidation-Reduction
  • Nuclear Chemistry
  • NGSS Alignments
  • 1D-Kinematics
  • Projectiles
  • Circular Motion
  • Magnetism and Electromagnetism
  • Graphing Practice
  • About the ACT
  • ACT Preparation
  • For Teachers
  • Other Resources
  • Newton's Laws of Motion
  • Work and Energy Packet
  • Static Electricity Review
  • Solutions Guide
  • Solutions Guide Digital Download
  • Motion in One Dimension
  • Work, Energy and Power
  • Frequently Asked Questions
  • Purchasing the Download
  • Purchasing the CD
  • Purchasing the Digital Download
  • About the NGSS Corner
  • NGSS Search
  • Force and Motion DCIs - High School
  • Energy DCIs - High School
  • Wave Applications DCIs - High School
  • Force and Motion PEs - High School
  • Energy PEs - High School
  • Wave Applications PEs - High School
  • Crosscutting Concepts
  • The Practices
  • Physics Topics
  • NGSS Corner: Activity List
  • NGSS Corner: Infographics
  • About the Toolkits
  • Position-Velocity-Acceleration
  • Position-Time Graphs
  • Velocity-Time Graphs
  • Newton's First Law
  • Newton's Second Law
  • Newton's Third Law
  • Terminal Velocity
  • Projectile Motion
  • Forces in 2 Dimensions
  • Impulse and Momentum Change
  • Momentum Conservation
  • Work-Energy Fundamentals
  • Work-Energy Relationship
  • Roller Coaster Physics
  • Satellite Motion
  • Electric Fields
  • Circuit Concepts
  • Series Circuits
  • Parallel Circuits
  • Describing-Waves
  • Wave Behavior Toolkit
  • Standing Wave Patterns
  • Resonating Air Columns
  • Wave Model of Light
  • Plane Mirrors
  • Curved Mirrors
  • Teacher Guide
  • Using Lab Notebooks
  • Current Electricity
  • Light Waves and Color
  • Reflection and Ray Model of Light
  • Refraction and Ray Model of Light
  • Classes (Legacy Version)
  • Teacher Resources
  • Subscriptions

steps for solving projectile motion problems

  • Newton's Laws
  • Einstein's Theory of Special Relativity
  • About Concept Checkers
  • School Pricing
  • Newton's Laws of Motion
  • Newton's First Law
  • Newton's Third Law
  • Horizontally Launched Projectile Problems
  • What is a Projectile?
  • Motion Characteristics of a Projectile
  • Horizontal and Vertical Velocity
  • Horizontal and Vertical Displacement
  • Initial Velocity Components
  • Non-Horizontally Launched Projectile Problems

steps for solving projectile motion problems

There are two basic types of projectile problems that we will discuss in this course. While the general principles are the same for each type of problem, the approach will vary due to the fact the problems differ in terms of their initial conditions. The two types of problems are:

A projectile is launched with an initial horizontal velocity from an elevated position and follows a parabolic path to the ground. Predictable unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight, and the horizontal distance of the projectile.

Examples of this type of problem are

  • A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.

A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

A projectile is launched at an angle to the horizontal and rises upwards to a peak while moving horizontally. Upon reaching the peak, the projectile falls with a motion that is symmetrical to its path upwards to the peak. Predictable unknowns include the time of flight, the horizontal range, and the height of the projectile when it is at its peak.

  • A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the football.
  • A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.

The second problem type will be the subject of the next part of Lesson 2 . In this part of Lesson 2, we will focus on the first type of problem - sometimes referred to as horizontally launched projectile problems. Three common kinematic equations that will be used for both type of problems include the following:

d = v i •t + 0.5*a*t 2 v f = v i + a•t v f 2  = v i 2  + 2*a•d  

Equations for the Horizontal Motion of a Projectile

The above equations work well for motion in one-dimension, but a projectile is usually moving in two dimensions - both horizontally and vertically. Since these two components of motion are independent of each other, two distinctly separate sets of equations are needed - one for the projectile's horizontal motion and one for its vertical motion. Thus, the three equations above are transformed into two sets of three equations. For the horizontal components of motion, the equations are

x = v i x •t + 0.5*a x *t 2

v f x  = v i x  + a x •t

v f x 2  = v i x 2  + 2*a x •x

Of these three equations, the top equation is the most commonly used. An application of projectile concepts to each of these equations would also lead one to conclude that any term with a x in it would cancel out of the equation since a x = 0 m/s/s . Once this cancellation of ax terms is performed, the only equation of usefulness is:

x = v i x •t

Equations for the Vertical Motion of a Projectile

For the vertical components of motion, the three equations are

y = v iy •t + 0.5*a y *t 2

v fy  = v iy  + a y •t

v fy 2  = v iy 2  + 2*a y •y

In each of the above equations, the vertical acceleration of a projectile is known to be -9.8 m/s/s (the acceleration of gravity). Furthermore, for the special case of the first type of problem (horizontally launched projectile problems), v iy = 0 m/s. Thus, any term with v iy in it will cancel out of the equation.

The two sets of three equations above are the kinematic equations that will be used to solve projectile motion problems.

Solving Projectile Problems

To illustrate the usefulness of the above equations in making predictions about the motion of a projectile, consider the solution to the following problem.

The solution of this problem begins by equating the known or given values with the symbols of the kinematic equations - x, y, v ix , v iy , a x , a y , and t. Because horizontal and vertical information is used separately, it is a wise idea to organized the given information in two columns - one column for horizontal information and one column for vertical information. In this case, the following information is either given or implied in the problem statement:

As indicated in the table, the unknown quantity is the horizontal displacement (and the time of flight) of the pool ball. The solution of the problem now requires the selection of an appropriate strategy for using the kinematic equations and the known information to solve for the unknown quantities. It will almost always be the case that such a strategy demands that one of the vertical equations be used to determine the time of flight of the projectile and then one of the horizontal equations be used to find the other unknown quantities (or vice versa - first use the horizontal and then the vertical equation). An organized listing of known quantities (as in the table above) provides cues for the selection of the strategy. For example, the table above reveals that there are three quantities known about the vertical motion of the pool ball. Since each equation has four variables in it, knowledge of three of the variables allows one to calculate a fourth variable. Thus, it would be reasonable that a vertical equation is used with the vertical values to determine time and then the horizontal equations be used to determine the horizontal displacement (x). The first vertical equation (y = v iy •t +0.5•a y •t 2 ) will allow for the determination of the time. Once the appropriate equation has been selected, the physics problem becomes transformed into an algebra problem. By substitution of known values, the equation takes the form of

Since the first term on the right side of the equation reduces to 0, the equation can be simplified to

If both sides of the equation are divided by -5.0 m/s/s, the equation becomes

By taking the square root of both sides of the equation, the time of flight can then be determined .

Once the time has been determined, a horizontal equation can be used to determine the horizontal displacement of the pool ball. Recall from the given information , v ix = 2.4 m/s and a x = 0 m/s/s. The first horizontal equation (x = v ix •t + 0.5•a x •t 2 ) can then be used to solve for "x." With the equation selected, the physics problem once more becomes transformed into an algebra problem. By substitution of known values, the equation takes the form of

Since the second term on the right side of the equation reduces to 0, the equation can then be simplified to

The answer to the stated problem is that the pool ball is in the air for 0.35 seconds and lands a horizontal distance of 0.84 m from the edge of the pool table.

The following procedure summarizes the above problem-solving approach.

  • Carefully read the problem and list known and unknown information in terms of the symbols of the kinematic equations. For convenience sake, make a table with horizontal information on one side and vertical information on the other side.
  • Identify the unknown quantity that the problem requests you to solve for.
  • Select either a horizontal or vertical equation to solve for the time of flight of the projectile.
  • With the time determined, use one of the other equations to solve for the unknown. (Usually, if a horizontal equation is used to solve for time, then a vertical equation can be used to solve for the final unknown quantity.)

One caution is in order. The sole reliance upon 4- and 5-step procedures to solve physics problems is always a dangerous approach. Physics problems are usually just that - problems! While problems can often be simplified by the use of short procedures as the one above, not all problems can be solved with the above procedure. While steps 1 and 2 above are critical to your success in solving horizontally launched projectile problems, there will always be a problem that doesn't fit the mold . Problem solving is not like cooking; it is not a mere matter of following a recipe. Rather, problem solving requires careful reading, a firm grasp of conceptual physics, critical thought and analysis, and lots of disciplined practice. Never divorce conceptual understanding and critical thinking from your approach to solving problems.

Check Your Understanding

Use y = v iy • t + 0.5 • a y • t 2 to solve for time; the time of flight is 2.12 seconds.

Now use x = v ix • t + 0.5 • a x • t 2 to solve for v ix

Note that a x is 0 m/s/s so the last term on the right side of the equation cancels. By substituting 35.0 m for x and 2.12 s for t, the v ix can be found to be 16.5 m/s.

We Would Like to Suggest ...

steps for solving projectile motion problems

  • Addition of Forces

Two-Dimensional Kinematics

Projectile motion, learning objectives.

By the end of this section, you will be able to:

  • Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory.
  • Determine the location and velocity of a projectile at different points in its trajectory.
  • Apply the principle of independence of motion to solve projectile motion problems.

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile , and its path is called its trajectory . The motion of falling objects, as covered in Problem-Solving Basics for One-Dimensional Kinematics, is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance  is negligible .

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction , where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x -axis and the vertical axis the y -axis. Figure 1 illustrates the notation for displacement, where s is defined to be the total displacement and x and y are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s , x , and y . (Note that in the last section we used the notation A to represent a vector with components A x and A y . If we continued this format, we would call displacement s with components s x and s y . However, to simplify the notation, we will simply represent the component vectors as x and y .)

Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x – and y -axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: a y  = – g  = –9.80 m/s 2 . (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, a x =0. Both accelerations are constant, so the kinematic equations can be used.

Review of Kinematic Equations (constant a )

A soccer player is kicking a soccer ball. The ball travels in a projectile motion and reaches a point whose vertical distance is y and horizontal distance is x. The displacement between the kicking point and the final point is s. The angle made by this displacement vector with x axis is theta.

Figure 1. The total displacement s of a soccer ball at a point along its path. The vector s has components x and y along the horizontal and vertical axes. Its magnitude is s, and it makes an angle θ with the horizontal.

Given these assumptions, the following steps are then used to analyze projectile motion:

Step 1.  Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so A x = A cos θ  and  A y  = A sin θ  are used. The magnitude of the components of displacement s along these axes are x and y. The magnitudes of the components of the velocity v are V x = V cos θ  and V y = v sin θ  where v is the magnitude of the velocity and θ is its direction, as shown in 2. Initial values are denoted with a subscript 0, as usual.

Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:

Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time t . The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below.

Step 4.  Recombine the two motions to find the total displacement s and velocity v. Because the x – and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing [latex]A=\sqrt{{{A}_{x}}^{2}+{{A}_{y}}^{2}}\\[/latex] and θ  = tan −1  ( A y / A x ) in the following form, where θ is the direction of the displacement s and θ v is the direction of the velocity v :

Total displacement and velocity

In part a the figure shows projectile motion of a ball with initial velocity of v zero at an angle of theta zero with the horizontal x axis. The horizontal component v x and the vertical component v y at various positions of ball in the projectile path is shown. In part b only the horizontal velocity component v sub x is shown whose magnitude is constant at various positions in the path. In part c only vertical velocity component v sub y is shown. The vertical velocity component v sub y is upwards till it reaches the maximum point and then its direction changes to downwards. In part d resultant v of horizontal velocity component v sub x and downward vertical velocity component v sub y is found which makes an angle theta with the horizontal x axis. The direction of resultant velocity v is towards south east.

Figure 2. (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because ax=0 and vx is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x – and y -motions are recombined to give the total velocity at any given point on the trajectory.

Example 1. A Fireworks Projectile Explodes High and Away

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0º above the horizontal, as illustrated in Figure 3. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which a x  = 0 and a y  = – g . We can then define x 0 and y 0 to be zero and solve for the desired quantities.

Solution for (a)

By “height” we mean the altitude or vertical position y above the starting point. The highest point in any trajectory, called the apex, is reached when v y =0. Since we know the initial and final velocities as well as the initial position, we use the following equation to find y :

The x y graph shows the trajectory of fireworks shell. The initial velocity of the shell v zero is at angle theta zero equal to seventy five degrees with the horizontal x axis. The fuse is set to explode the shell at the highest point of the trajectory which is at a height h equal to two hundred thirty three meters and at a horizontal distance x equal to one hundred twenty five meters from the origin.

Figure 3. The trajectory of a fireworks shell. The fuse is set to explode the shell at the highest point in its trajectory, which is found to be at a height of 233 m and 125 m away horizontally.

Because y 0 and v y are both zero, the equation simplifies to

Solving for y gives

Now we must find v 0 y , the component of the initial velocity in the y -direction. It is given by v 0y = v 0 sin  θ , where v 0 y is the initial velocity of 70.0 m/s, and θ 0  = 75.0º is the initial angle. Thus,

v Oy = v 0 sin  θ 0 = (70.0 m/s)(sin 75º) = 67.6 m/s

[latex]y=\frac{\left(67.6\text{ m/s}\right)^{2}}{2\left(9.80\text{ m/s}^{2}\right)}\\[/latex] ,

Discussion for (a)

Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height.

Solution for (b)

As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use [latex]y={y}_{0}+\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t\\[/latex]. Because y 0 is zero, this equation reduces to simply

[latex]y=\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t\\[/latex].

Note that the final vertical velocity, v y , at the highest point is zero. Thus,

[latex]\begin{array}{lll}t& =& \frac{2y}{\left({v}_{0y}+{v}_{y}\right)}=\frac{2\left(\text{233 m}\right)}{\left(\text{67.6 m/s}\right)}\\ & =& 6.90\text{ s}\end{array}\\[/latex].

Discussion for (b)

This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. (Another way of finding the time is by using [latex]y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\text{gt}}^{2}\\[/latex], and solving the quadratic equation for t .)

Solution for (c)

Because air resistance is negligible, a x =0 and the horizontal velocity is constant, as discussed above. The horizontal displacement is horizontal velocity multiplied by time as given by x  =  x 0  +  v x t , where x 0 is equal to zero:

x  =  v x t ,

where v x is the x -component of the velocity, which is given by v x = v 0 cos θ 0  Now,

v x  = v 0  cos  θ 0 = (70.0 m/s)(cos 75º) = 18.1 m/s

The time t for both motions is the same, and so x is

x = (18.1 m/s)(6.90 s) = 125 m.

Discussion for (c)

The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below.

In solving part (a) of the preceding example, the expression we found for y is valid for any projectile motion where air resistance is negligible. Call the maximum height y = h ; then,

[latex]h=\frac{{{v}_{0y}}^{2}}{2g}\\[/latex].

This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity.

Defining a Coordinate System

Example 2. calculating projectile motion: hot rock projectile.

Kilauea in Hawaii is the world’s most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 25.0 m/s and at an angle 35.0º above the horizontal, as shown in Figure 4. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and direction of the rock’s velocity at impact?

The trajectory of a rock ejected from a volcano is shown. The initial velocity of rock v zero is equal to twenty five meters per second and it makes an angle of thirty five degrees with the horizontal x axis. The figure shows rock falling down a height of twenty meters below the volcano level. The velocity at this point is v which makes an angle of theta with horizontal x axis. The direction of v is south east.

Figure 4. The trajectory of a rock ejected from the Kilauea volcano.

Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. We will solve for t first. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be recombined to obtain v and θ v at the final time t determined in the first part of the example.

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

[latex]y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\text{gt}}^{2}\\[/latex].

If we take the initial position y 0 to be zero, then the final position is y  = −20.0 m. Now the initial vertical velocity is the vertical component of the initial velocity, found from  v Oy = v 0 sin  θ 0  = (25.0 m/s)(sin 35.0º) = 14.3 m/s. Substituting known values yields

Rearranging terms gives a quadratic equation in t :

This expression is a quadratic equation of the form at 2 + bt + c = 0 , where the constants are a  = 4.90 , b  = –14.3 , and c  = –20.0. Its solutions are given by the quadratic formula:

[latex]t=\frac{-bpm \sqrt{{b}^{2}-4\text{ac}}}{\text{2}\text{a}}\\[/latex]

This equation yields two solutions: t  = 3.96 and t  = –1.03. (It is left as an exercise for the reader to verify these solutions.) The time is t = 3.96 s or -1.03 s. The negative value of time implies an event before the start of motion, and so we discard it. Thus,

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

From the information now in hand, we can find the final horizontal and vertical velocities v x and v y and combine them to find the total velocity v and the angle θ 0 it makes with the horizontal. Of course, v x is constant so we can solve for it at any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle. Therefore:

  v x  = v 0  cos  θ 0  = (25.0 m/s)(cos 35º) = 20.5 m/s

The final vertical velocity is given by the following equation:

[latex]{v}_{y}={v}_{0y}\text{gt}\\[/latex],

where v 0y was found in part (a) to be 14.3 m/s. Thus,

To find the magnitude of the final velocity v we combine its perpendicular components, using the following equation:

[latex]v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}=\sqrt{({20.5}\text{ m/s})^{2}+{({-24.5}\text{ m/s})^{2}}}\\[/latex]

which gives

The direction θ v is found from the equation:

The negative angle means that the velocity is 50.1º below the horizontal. This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See Figure 4.)

Part a of the figure shows three different trajectories of projectiles on level ground. In each case the projectiles makes an angle of forty five degrees with the horizontal axis. The first projectile of initial velocity thirty meters per second travels a horizontal distance of R equal to ninety one point eight meters. The second projectile of initial velocity forty meters per second travels a horizontal distance of R equal to one hundred sixty three meters. The third projectile of initial velocity fifty meters per second travels a horizontal distance of R equal to two hundred fifty five meters.

Figure 5. Trajectories of projectiles on level ground. (a) The greater the initial speed v0, the greater the range for a given initial angle. (b) The effect of initial angle θ 0  on the range of a projectile with a given initial speed. Note that the range is the same for 15º and 75º, although the maximum heights of those paths are different.

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed v 0 , the greater the range, as shown in Figure 5(a). The initial angle θ 0 also has a dramatic effect on the range, as illustrated in Figure 5(b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with θ 0  = 45º. This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately 38º. Interestingly, for every initial angle except 45º, there are two angles that give the same range—the sum of those angles is 90º. The range also depends on the value of the acceleration of gravity g . The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range R of a projectile on level ground for which air resistance is negligible is given by

[latex]R=\frac{{{v}_{0}}^{2}\sin{2\theta }_{0}}{g}\\[/latex],

where v 0 is the initial speed and θ 0 is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described. When we speak of the range of a projectile on level ground, we assume that R is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See Figure 6.) If the initial speed is great enough, the projectile goes into orbit. This is called escape velocity. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text. Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Addition of Velocities , we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic.

A figure of the Earth is shown and on top of it a very high tower is placed. A projectile satellite is launched from this very high tower with initial velocity of v zero in the horizontal direction. Several trajectories are shown with increasing range. A circular trajectory is shown indicating the satellite achieved its orbit and it is revolving around the Earth.

Figure 6. Projectile to satellite. In each case shown here, a projectile is launched from a very high tower to avoid air resistance. With increasing initial speed, the range increases and becomes longer than it would be on level ground because the Earth curves away underneath its path. With a large enough initial speed, orbit is achieved.

PhET Explorations: Projectile Motion

Click to run the simulation.

Section Summary

  • Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity.
  • To solve projectile motion problems, perform the following steps:

1. Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position s are given by the quantities x  and y , and the components of the velocity v are given by v x =   v cos θ  and v y = v sin θ , where v  is the magnitude of the velocity and  θ  is its direction.

2. Analyze the motion of the projectile in the horizontal direction using the following equations:

Horizontal Motion ( a x = 0)

3. Analyze the motion of the projectile in the vertical direction using the following equations:

Vertical Motion (assuming positive is up a y = -g = -9.8 m/s 2 )

4. Recombine the horizontal and vertical components of location and/or velocity using the following equations:

  • The maximum height  h of a projectile launched with initial vertical velocity v 0y is given by [latex]h=\frac{{{v}_{0y}}^{2}}{2g}\\[/latex].
  • The maximum horizontal distance traveled by a projectile is called the range . The range R of a projectile on level ground launched at an angle  θ 0   above the horizontal with initial speed v 0 is given by [latex]R=\frac{{{{v}_{0}}^{2}}\text{\sin}{2\theta }_{0}}{g}\\[/latex].

Conceptual Questions

1. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0º nor 90º): (a) Is the velocity ever zero? (b) When is the velocity a minimum? A maximum? (c) Can the velocity ever be the same as the initial velocity at a time other than at t = 0? (d) Can the speed ever be the same as the initial speed at a time other than at  t = 0?

2. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0º nor 90º): (a) Is the acceleration ever zero? (b) Is the acceleration ever in the same direction as a component of velocity? (c) Is the acceleration ever opposite in direction to a component of velocity?

3. For a fixed initial speed, the range of a projectile is determined by the angle at which it is fired. For all but the maximum, there are two angles that give the same range. Considering factors that might affect the ability of an archer to hit a target, such as wind, explain why the smaller angle (closer to the horizontal) is preferable. When would it be necessary for the archer to use the larger angle? Why does the punter in a football game use the higher trajectory?

4. During a lecture demonstration, a professor places two coins on the edge of a table. She then flicks one of the coins horizontally off the table, simultaneously nudging the other over the edge. Describe the subsequent motion of the two coins, in particular discussing whether they hit the floor at the same time.

Problems & Exercises

1. A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0º above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the x and y distances from where the projectile was launched to where it lands?

2. A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction. (a) At what speed does the ball hit the ground? (b) For how long does the ball remain in the air? (c)What maximum height is attained by the ball?

3. A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

4. (a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32º ramp at a speed of 40.0 m/s (144 km/h). How many buses can he clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 20.0 m long? (b) Discuss what your answer implies about the margin of error in this act—that is, consider how much greater the range is than the horizontal distance he must travel to miss the end of the last bus. (Neglect air resistance.)

5. An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

6. A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand. (a) At what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used? (b) What other angle gives the same range, and why would it not be used? (c) How long did this pass take?

7. Verify the ranges for the projectiles in Figure 5 (a) for θ = 45º and the given initial velocities.

8. Verify the ranges shown for the projectiles in Figure 5(b) for an initial velocity of 50 m/s at the given initial angles.

9. The cannon on a battleship can fire a shell a maximum distance of 32.0 km. (a) Calculate the initial velocity of the shell. (b) What maximum height does it reach? (At its highest, the shell is above 60% of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.) (c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is 6.37 × 10 3 . How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?

10. An arrow is shot from a height of 1.5 m toward a cliff of height  H . It is shot with a velocity of 30 m/s at an angle of 60º above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow’s impact speed just before hitting the cliff?

11. In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

12. The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions.

13. Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle  θ  below the horizontal. The service line is 11.9 m from the net, which is 0.91 m high. What is the angle  θ such that the ball just crosses the net? Will the ball land in the service box, whose out line is 6.40 m from the net?

14. A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. (a) If the ball is thrown at an angle of 25º relative to the ground and is caught at the same height as it is released, what is its initial speed relative to the ground? (b) How long does it take to get to the receiver? (c) What is its maximum height above its point of release?

15. Gun sights are adjusted to aim high to compensate for the effect of gravity, effectively making the gun accurate only for a specific range. (a) If a gun is sighted to hit targets that are at the same height as the gun and 100.0 m away, how low will the bullet hit if aimed directly at a target 150.0 m away? The muzzle velocity of the bullet is 275 m/s. (b) Discuss qualitatively how a larger muzzle velocity would affect this problem and what would be the effect of air resistance.

16. An eagle is flying horizontally at a speed of 3.00 m/s when the fish in her talons wiggles loose and falls into the lake 5.00 m below. Calculate the velocity of the fish relative to the water when it hits the water.

17. An owl is carrying a mouse to the chicks in its nest. Its position at that time is 4.00 m west and 12.0 m above the center of the 30.0 cm diameter nest. The owl is flying east at 3.50 m/s at an angle 30.0º below the horizontal when it accidentally drops the mouse. Is the owl lucky enough to have the mouse hit the nest? To answer this question, calculate the horizontal position of the mouse when it has fallen 12.0 m.

18. Suppose a soccer player kicks the ball from a distance 30 m toward the goal. Find the initial speed of the ball if it just passes over the goal, 2.4 m above the ground, given the initial direction to be 40º above the horizontal.

19. Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s.

20. The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 7.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved in solving projectile motion problems.

21. In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77 m. What was the initial speed of the shot if he released it at a height of 2.10 m and threw it at an angle of 38.0º above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at 45º  when air resistance is neglected, the actual angle to achieve maximum range is smaller; thus, 38º  will give a longer range than 45º  in the shot put.)

22. A basketball player is running at 5.00 m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity. (a) What vertical velocity does he need to rise 0.750 m above the floor? (b) How far from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket?

23. A football player punts the ball at a 45º angle. Without an effect from the wind, the ball would travel 60.0 m horizontally. (a) What is the initial speed of the ball? (b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.50 m/s. What distance does the ball travel horizontally?

24. Prove that the trajectory of a projectile is parabolic, having the form [latex]y=\text{ax}+{\text{bx}}^{2}\\[/latex]. To obtain this expression, solve the equation [latex]x={v}_{0x}t\\[/latex] for t  and substitute it into the expression for [latex]y={v}_{0y}t-\left(1/2\right){\text{gt}}^{2}\\[/latex]. (These equations describe the x  and y positions of a projectile that starts at the origin.) You should obtain an equation of the form [latex]y=\text{ax}+{\text{bx}}^{2}\\[/latex] where a and b are constants.

25. Derive [latex]R=\frac{{{v}_{0}}^{2}\text{\sin}{2\theta }_{0}}{g}\\[/latex] for the range of a projectile on level ground by finding the time t at which y becomes zero and substituting this value of t  into the expression for x – x 0 , noting that R = x – x 0 .

26.  Unreasonable Results (a) Find the maximum range of a super cannon that has a muzzle velocity of 4.0 km/s. (b) What is unreasonable about the range you found? (c) Is the premise unreasonable or is the available equation inapplicable? Explain your answer. (d) If such a muzzle velocity could be obtained, discuss the effects of air resistance, thinning air with altitude, and the curvature of the Earth on the range of the super cannon.

27. Construct Your Own Problem Consider a ball tossed over a fence. Construct a problem in which you calculate the ball’s needed initial velocity to just clear the fence. Among the things to determine are; the height of the fence, the distance to the fence from the point of release of the ball, and the height at which the ball is released. You should also consider whether it is possible to choose the initial speed for the ball and just calculate the angle at which it is thrown. Also examine the possibility of multiple solutions given the distances and heights you have chosen.

Selected Solutions to Problems & Exercises

1.  x = 1.30 m × 10 2 , y = 30.9 m

3. (a) 3.50 s (b) 28.6 m/s (c) 34.3 m/s (d) 44.7 m/s, 50.2º below horizontal

5. (a) 18.4º (b) The arrow will go over the branch.

7. [latex]R=\frac{{{{v}_{0}}}^{}}{\sin{2\theta }_{0}g}\\[/latex]

For θ = 45º, [latex]R=\frac{{{{v}_{0}}}^{2}}{g}\\[/latex]

R = 91.9 m for v 0  = 30 m/s; R = 163 m for v 0 ; R = 255 m for v 0 = 50 m/s

9. (a) 560 m/s (b) 800 × 10 3 m  (c) 80.0 m. This error is not significant because it is only 1% of the answer in part (b).

11. 1.50 m, assuming launch angle of 45º

13.  θ  =6.1º. Yes, the ball lands at 5.3 m from the net

15. (a) −0.486 m (b) The larger the muzzle velocity, the smaller the deviation in the vertical direction, because the time of flight would be smaller. Air resistance would have the effect of decreasing the time of flight, therefore increasing the vertical deviation.

17. 4.23 m. No, the owl is not lucky; he misses the nest.

19. No, the maximum range (neglecting air resistance) is about 92 m.

21. 15.0 m/s

23. (a) 24.2 m/s (b) The ball travels a total of 57.4 m with the brief gust of wind.

25. [latex]y-{y}_{0}=0={v}_{0y}t-\frac{1}{2}{gt}^{2}=\left({v}_{0}\sin\theta\right)t-\frac{1}{2}{gt}^{2}\\[/latex] ,

so that [latex]t=\frac{2\left({v}_{0}\sin\theta \right)}{g}\\[/latex]

[latex]x-{x}_{0}={v}_{0x}t=\left({v}_{0}\cos\theta \right)t=R\\[/latex], and substituting for t  gives:

[latex]R={v}_{0}\cos\theta \left(\frac{{2v}_{0}\sin\theta}{g}\right)=\frac{{{2v}_{0}}^{2}\sin\theta \cos\theta }{g}\\[/latex]

since [latex]2\sin\theta \cos\theta =\sin 2\theta\\[/latex], the range is:

[latex]R=\frac{{{v}_{0}}^{2}\sin 2\theta }{g}\\[/latex].

  • College Physics. Authored by : OpenStax College. Located at : http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a/College_Physics . License : CC BY: Attribution . License Terms : Located at License
  • PhET Interactive Simulations . Authored by : University of Colorado Boulder . Located at : http://phet.colorado.edu . License : CC BY: Attribution

Library homepage

  • school Campus Bookshelves
  • menu_book Bookshelves
  • perm_media Learning Objects
  • login Login
  • how_to_reg Request Instructor Account
  • hub Instructor Commons
  • Download Page (PDF)
  • Download Full Book (PDF)
  • Periodic Table
  • Physics Constants
  • Scientific Calculator
  • Reference & Cite
  • Tools expand_more
  • Readability

selected template will load here

This action is not available.

Physics LibreTexts

4.4: Projectile Motion

  • Last updated
  • Save as PDF
  • Page ID 45974

Learning Objectives

  • Use one-dimensional motion in perpendicular directions to analyze projectile motion.
  • Calculate the range, time of flight, and maximum height of a projectile that is launched and impacts a flat, horizontal surface.
  • Find the time of flight and impact velocity of a projectile that lands at a different height from that of launch.
  • Calculate the trajectory of a projectile.

Projectile motion is the motion of an object thrown or projected into the air, subject only to acceleration as a result of gravity. The applications of projectile motion in physics and engineering are numerous. Some examples include meteors as they enter Earth’s atmosphere, fireworks, and the motion of any ball in sports. Such objects are called projectiles and their path is called a trajectory . The motion of falling objects as discussed in Motion Along a Straight Line is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, and our treatment neglects the effects of air resistance.

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. We discussed this fact in Displacement and Velocity Vectors , where we saw that vertical and horizontal motions are independent. The key to analyzing two-dimensional projectile motion is to break it into two motions: one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible because acceleration resulting from gravity is vertical; thus, there is no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x-axis and the vertical axis the y-axis. It is not required that we use this choice of axes; it is simply convenient in the case of gravitational acceleration. In other cases we may choose a different set of axes. Figure \(\PageIndex{1}\) illustrates the notation for displacement, where we define \(\vec{s}\) to be the total displacement, and \(\vec{x}\) and \(\vec{y}\) are its component vectors along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y.

An illustration of a soccer player kicking a ball. The soccer player’s foot is at the origin of an x y coordinate system. The trajectory of the soccer ball and its location at 6 instants in time are shown. The trajectory is a parabola. The vector s is the displacement from the origin to the final position of the soccer ball. Vector s and its x and y components form a right triangle, with s as the hypotenuse and an angle phi between the x axis and s.

To describe projectile motion completely, we must include velocity and acceleration, as well as displacement. We must find their components along the x- and y-axes. Let’s assume all forces except gravity (such as air resistance and friction, for example) are negligible. Defining the positive direction to be upward, the components of acceleration are then very simple:

\[a_{y} = −g = −9.8\; m/s^{2} (− 32\; ft/s^{2}) \ldotp\]

Because gravity is vertical, a x = 0. If a x = 0, this means the initial velocity in the x direction is equal to the final velocity in the x direction, or v x = v 0x . With these conditions on acceleration and velocity, we can write the kinematic Equation 4.11 through Equation 4.18 for motion in a uniform gravitational field, including the rest of the kinematic equations for a constant acceleration from Motion with Constant Acceleration. The kinematic equations for motion in a uniform gravitational field become kinematic equations with a y = −g, a x = 0:

Horizontal Motion

\[v_{0x} = v_{x}, \quad x = x_{0} + v_{x} t \label{4.19}\]

Vertical Motion

\[y = y_{0} + \frac{1}{2} (v_{0y} + v_{y})t \label{4.20}\]

\[v_{y} = v_{0y} - gt \label{4.21}\]

\[y = y_{0} + v_{0y} t - \frac{1}{2} g t^{2} \label{4.22}\]

\[v_{y}^{2}= v_{0y}^{2} + 2g(y − y_{0}) \label{4.23}\]

Using this set of equations, we can analyze projectile motion, keeping in mind some important points.

Problem-Solving Strategy: Projectile Motion

  • Resolve the motion into horizontal and vertical components along the x- and y-axes. The magnitudes of the components of displacement \(\vec{s}\) along these axes are x and y. The magnitudes of the components of velocity \(\vec{v}\) are v x = vcos\(\theta\) and v y = vsin\(\theta\), where v is the magnitude of the velocity and \(\theta\) is its direction relative to the horizontal, as shown in Figure \(\PageIndex{2}\).
  • Treat the motion as two independent one-dimensional motions: one horizontal and the other vertical. Use the kinematic equations for horizontal and vertical motion presented earlier.
  • Solve for the unknowns in the two separate motions: one horizontal and one vertical. Note that the only common variable between the motions is time t. The problem-solving procedures here are the same as those for one-dimensional kinematics and are illustrated in the following solved examples.
  • Recombine quantities in the horizontal and vertical directions to find the total displacement \(\vec{s}\) and velocity \(\vec{v}\). Solve for the magnitude and direction of the displacement and velocity using $$s = \sqrt{x^{2} + y^{2}} \ldotp \quad \phi = \tan^{-1} \left(\dfrac{y}{x}\right), \quad v = \sqrt{v_{x}^{2} + v_{y}^{2}} \ldotp$$where \(\phi\) is the direction of the displacement \(\vec{s}\).

Figure a shows the locations and velocities of a projectile on an x y coordinate system at 10 instants in time. When the projectile is at the origin, it has a velocity v sub 0 y which makes an angle theta sub 0 with the horizontal. The velocity is shown as a dark blue arrow, and its x and y components are shown as light blue arrow. The projectile’s position follows a downward-opening parabola, moving up to a maximum height, then back to y = 0, and continuing below the x axis The velocity, V, at each time makes an angle theta which changes in time, and has x component V sub x and y component v sub y. The x component of the velocity V sub x is the same at all times. The y component v sub y points up but gets smaller, until the projectile reaches the maximum height, where the velocity is horizontal and has no y component. After the maximum height, the velocity has a y component pointing down and growing larger. As the projectile reaches the same elevation on the way down as it had on the way up, its velocity is below the horizontal by the same angle theta as it was above the horizontal on the way up. In particular, when it comes back to y = 0 on the way down, the angle between the vector v and the horizontal is minus s=theta sub zero and the y component of the velocity is minus v sub 0 y. The last position shown is below the x axis, and the y component of the velocity is larger than it was initially. The graph clearly shows that the horizontal distances travelled in each of the time intervals are equal, while the vertical distances decrease on the way up and increase on the way down. Figure b shows the horizontal component, constant velocity. The horizontal positions and x components of the velocity of the projectile are shown along a horizontal line. The positions are evenly spaced, and the x components of the velocities are all the same, and point to the right. Figure c shows the vertical component, constant acceleration. The vertical positions and y components of the velocity of the projectile are shown along a vertical line. The positions are get closer together on the way up, then further apart on the way down. The y components of the velocities initially point up, decreasing in magnitude until there is no y component to the velocity at the maximum height. After the maximum height, the y components of the velocities point down and increase in magnitude. Figure d shows that putting the horizontal and vertical components of figures b and c together gives the total velocity at a point. The velocity V has an x component of V sub x, has y component of V sub y, and makes an angle of theta with the horizontal. In the example shown, the velocity has a downward y component.

Example 4.7: A Fireworks Projectile Explodes high and away

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0° above the horizontal, as illustrated in Figure \(\PageIndex{3}\). The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passes between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? (d) What is the total displacement from the point of launch to the highest point?

The trajectory of a fireworks shell from its launch to its highest point is shown as the left half of a downward-opening parabola in a graph of y as a function of x. The maximum height is h = 233 meters and its x displacement at that time is x = 125 meters. The initial velocity vector v sub 0 is up and to the right, tangent to the trajectory curve, and makes an angle of theta sub 0 equal to 75 degrees.

The motion can be broken into horizontal and vertical motions in which a x = 0 and a y = −g. We can then define x 0 and y 0 to be zero and solve for the desired quantities.

  • By “height” we mean the altitude or vertical position y above the starting point. The highest point in any trajectory, called the apex, is reached when v y = 0. Since we know the initial and final velocities, as well as the initial position, we use the following equation to find y: $$v_{y}^{2} = v_{0y}^{2} - 2g(y - y_{0}) \ldotp$$Because y 0 and v y are both zero, the equation simplifies to $$0 = v_{0y}^{2} - 2gy \ldotp$$Solving for y gives $$y = \frac{v_{0y}^{2}}{2g} \ldotp$$Now we must find v 0y , the component of the initial velocity in the y direction. It is given by v 0y = v 0 sin\(\theta_{0}\), where v 0 is the initial velocity of 70.0 m/s and \(\theta_{0}\) = 75° is the initial angle. Thus $$v_{0y} = v_{0} \sin \theta = (70.0\; m/s) \sin 75^{o} = 67.6\; m/s$$and y is $$y = \frac{(67.6\; m/s)^{2}}{2(9.80\; m/s^{2})} \ldotp$$Thus, we have $$y = 233\; m \ldotp$$Note that because up is positive, the initial vertical velocity is positive, as is the maximum height, but the acceleration resulting from gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6-m/s initial vertical component of velocity reaches a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, so the initial velocity would have to be somewhat larger than that given to reach the same height.
  • As in many physics problems, there is more than one way to solve for the time the projectile reaches its highest point. In this case, the easiest method is to use v y = v 0y − gt. Because v y = 0 at the apex, this equation reduces $$0 = v_{0y} - gt$$or $$t = \frac{v_{0y}}{g} = \frac{67.6\; m/s}{9.80\; m/s^{2}} = 6.90\; s \ldotp$$This time is also reasonable for large fireworks. If you are able to see the launch of fireworks, notice that several seconds pass before the shell explodes. Another way of finding the time is by using y = y 0 + \(\frac{1}{2}\)(v 0y + v y )t. This is left for you as an exercise to complete.
  • Because air resistance is negligible, a x = 0 and the horizontal velocity is constant, as discussed earlier. The horizontal displacement is the horizontal velocity multiplied by time as given by x = x 0 + v x t, where x 0 is equal to zero. Thus, $$x = v_{x} t,$$where v x is the x-component of the velocity, which is given by $$v_{x} = v_{0} \cos \theta = (70.0\; m/s) \cos 75^{o} = 18.1\; m/s \ldotp$$Time t for both motions is the same, so x is $$x = (18.1\; m/s)(6.90\; s) = 125\; m \ldotp$$Horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. When the shell explodes, air resistance has a major effect, and many fragments land directly below.
  • The horizontal and vertical components of the displacement were just calculated, so all that is needed here is to find the magnitude and direction of the displacement at the highest point: $$\vec{s} = 125 \hat{i} + 233 \hat{j}$$$$|\vec{s}| = \sqrt{125^{2} + 233^{2}} = 264\; m$$$$\theta = \tan^{-1} \left(\dfrac{233}{125}\right) = 61.8^{o} \ldotp$$Note that the angle for the displacement vector is less than the initial angle of launch. To see why this is, review Figure \(\PageIndex{1}\), which shows the curvature of the trajectory toward the ground level. When solving Example 4.7(a), the expression we found for y is valid for any projectile motion when air resistance is negligible. Call the maximum height y = h. Then, $$h = \frac{v_{0y}^{2}}{2g} \ldotp$$This equation defines the maximum height of a projectile above its launch position and it depends only on the vertical component of the initial velocity.

Exercise 4.3

A rock is thrown horizontally off a cliff 100.0 m high with a velocity of 15.0 m/s. (a) Define the origin of the coordinate system. (b) Which equation describes the horizontal motion? (c) Which equations describe the vertical motion? (d) What is the rock’s velocity at the point of impact?

Example 4.8: Calculating projectile motion- Tennis Player

A tennis player wins a match at Arthur Ashe stadium and hits a ball into the stands at 30 m/s and at an angle 45° above the horizontal (Figure \(\PageIndex{4}\)). On its way down, the ball is caught by a spectator 10 m above the point where the ball was hit. (a) Calculate the time it takes the tennis ball to reach the spectator. (b) What are the magnitude and direction of the ball’s velocity at impact?

An illustration of a tennis ball launched into the stands. The player is to the left of the stands and hits the ball up and to the right at an angle of theta equal to 45 degrees and velocity of v sub 0 equal to 30 meters per second. The ball reaches a spectator who is seated 10 meters above the initial height of the ball.

Again, resolving this two-dimensional motion into two independent one-dimensional motions allows us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. Thus, we solve for t first. While the ball is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, we recombine the vertical and horizontal results to obtain \(\vec{v}\) at final time t, determined in the first part of the example.

  • While the ball is in the air, it rises and then falls to a final position 10.0 m higher than its starting altitude. We can find the time for this by using Equation \ref{4.22}: $$y = y_{0} + v_{0y}t - \frac{1}{2} gt^{2} \ldotp$$If we take the initial position y 0 to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity: $$v_{0y} = v_{0} \sin \theta_{0} = (30.0\; m/s) \sin 45^{o} = 21.2\; m/s \ldotp$$Substituting into Equation \ref{4.22} for y gives us $$10.0\; m = (21.2\; m/s)t − (4.90\; m/s^{2})t^{2} \ldotp$$Rearranging terms gives a quadratic equation in t: $$(4.90\; m/s^{2})t^{2} − (21.2\; m/s)t + 10.0\; m = 0 \ldotp$$Use of the quadratic formula yields t = 3.79 s and t = 0.54 s. Since the ball is at a height of 10 m at two times during its trajectory—once on the way up and once on the way down—we take the longer solution for the time it takes the ball to reach the spectator: $$t = 3.79\; s \ldotp$$The time for projectile motion is determined completely by the vertical motion. Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m above its starting altitude spends 3.79 s in the air.
  • We can find the final horizontal and vertical velocities v x and v y with the use of the result from (a). Then, we can combine them to find the magnitude of the total velocity vector \(\vec{v}\) and the angle \(\theta\) it makes with the horizontal. Since v x is constant, we can solve for it at any horizontal location. We choose the starting point because we know both the initial velocity and the initial angle. Therefore, $$v_{x} = v_{0} \cos \theta_{0} = (30\; m/s) \cos 45^{o} = 21.2\; m/s \ldotp$$The final vertical velocity is given by Equation \ref{4.21}: $$v_{y} = v_{0y} − gt \ldotp$$Since \(v_{0y}\) was found in part (a) to be 21.2 m/s, we have $$v_{y} = 21.2\; m/s − (9.8\; m/s^{2})(3.79 s) = −15.9\; m/s \ldotp$$The magnitude of the final velocity \(\vec{v}\) is $$v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(21.2\; m/s)^{2} + (-15.9\; m/s)^{2}} = 26.5\; m/s \ldotp$$The direction \(\theta_{v}\) is found using the inverse tangent: $$\theta_{v} = \tan^{-1} \left(\dfrac{v_{y}}{v_{x}}\right) = \tan^{-1} \left(\dfrac{21.2}{-15.9}\right) = -53.1^{o} \ldotp$$

Significance

  • As mentioned earlier, the time for projectile motion is determined completely by the vertical motion. Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m above its starting altitude spends 3.79 s in the air.
  • The negative angle means the velocity is 53.1° below the horizontal at the point of impact. This result is consistent with the fact that the ball is impacting at a point on the other side of the apex of the trajectory and therefore has a negative y component of the velocity. The magnitude of the velocity is less than the magnitude of the initial velocity we expect since it is impacting 10.0 m above the launch elevation.

Time of Flight, Trajectory, and Range

Of interest are the time of flight, trajectory, and range for a projectile launched on a flat horizontal surface and impacting on the same surface. In this case, kinematic equations give useful expressions for these quantities, which are derived in the following sections.

Time of flight

We can solve for the time of flight of a projectile that is both launched and impacts on a flat horizontal surface by performing some manipulations of the kinematic equations. We note the position and displacement in y must be zero at launch and at impact on an even surface. Thus, we set the displacement in y equal to zero and find

\[y − y_{0} = v_{0y} t − \frac{1}{2} gt^{2} = (v_{0} \sin \theta_{0})t − \frac{1}{2} gt^{2} = 0 \ldotp\]

Factoring, we have

\[t \left(v_{0} \sin \theta_{0} - \dfrac{gt}{2}\right) = 0 \ldotp\]

Solving for t gives us

\[T_{tof} = \frac{2(v_{0} \sin \theta_{0})}{g} \ldotp \label{4.24}\]

This is the time of flight for a projectile both launched and impacting on a flat horizontal surface. Equation \ref{4.24} does not apply when the projectile lands at a different elevation than it was launched, as we saw in Example 4.8 of the tennis player hitting the ball into the stands. The other solution, t = 0, corresponds to the time at launch. The time of flight is linearly proportional to the initial velocity in the y direction and inversely proportional to g. Thus, on the Moon, where gravity is one-sixth that of Earth, a projectile launched with the same velocity as on Earth would be airborne six times as long.

The trajectory of a projectile can be found by eliminating the time variable t from the kinematic equations for arbitrary t and solving for y(x). We take x 0 = y 0 = 0 so the projectile is launched from the origin. The kinematic equation for x gives

\[x = v_{0x}t \Rightarrow t = \frac{x}{v_{0x}} = \frac{x}{v_{0} \cos \theta_{0}} \ldotp\]

Substituting the expression for t into the equation for the position y = (v 0 sin \(\theta_{0}\))t − \(\frac{1}{2}\) gt 2 gives

\[y = (v_{0} \sin \theta_{0}) \left(\dfrac{x}{v_{0} \cos \theta_{0}}\right) - \frac{1}{2} g \left(\dfrac{x}{v_{0} \cos \theta_{0}}\right)^{2} \ldotp\]

Rearranging terms, we have

\[y = (\tan \theta_{0})x - \Big[ \frac{g}{2(v_{0} \cos \theta_{0})^{2}} \Big] x^{2} \ldotp \label{4.25}\]

This trajectory equation is of the form y = ax + bx 2 , which is an equation of a parabola with coefficients

\[a = \tan \theta_{0}, \quad b = - \frac{g}{2(v_{0} \cos \theta_{0})^{2}} \ldotp\]

From the trajectory equation we can also find the range , or the horizontal distance traveled by the projectile. Factoring Equation \ref{4.25}, we have

\[y = x \Big[ \tan \theta_{0} - \frac{g}{2(v_{0} \cos \theta_{0})^{2}} x \Big] \ldotp\]

The position y is zero for both the launch point and the impact point, since we are again considering only a flat horizontal surface. Setting y = 0 in this equation gives solutions x = 0, corresponding to the launch point, and

\[x = \frac{2 v_{0}^{2} \sin \theta_{0} \cos \theta_{0}}{g} ,\]

corresponding to the impact point. Using the trigonometric identity 2sin \(\theta\)cos\(\theta\) = sin2\(\theta\) and setting x = R for range, we find

\[R = \frac{v_{0}^{2} \sin 2 \theta_{0}}{g} \ldotp \label{4.26}\]

Note particularly that Equation \ref{4.26} is valid only for launch and impact on a horizontal surface. We see the range is directly proportional to the square of the initial speed v 0 and sin 2 \(\theta_{0}\), and it is inversely proportional to the acceleration of gravity. Thus, on the Moon, the range would be six times greater than on Earth for the same initial velocity. Furthermore, we see from the factor sin 2 \(\theta_{0}\) that the range is maximum at 45°. These results are shown in Figure \(\PageIndex{5}\). In (a) we see that the greater the initial velocity, the greater the range. In (b), we see that the range is maximum at 45°. This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is somewhat smaller. It is interesting that the same range is found for two initial launch angles that sum to 90°. The projectile launched with the smaller angle has a lower apex than the higher angle, but they both have the same range.

Figure a shows the trajectories of projectiles launched at the same initial 45 degree angle above the horizontal and different initial velocities. The trajectories are shown from launch to landing back at the initial elevation. In orange is the trajectory for 30 meters per second, giving a range R (distance from launch to landing) of 91.8 m. In purple is the trajectory for 40 meters per second, giving a range R of 163 m. In blue is the trajectory for 50 meters per second, giving a range R of 255 m. The maximum height of the projectile increases with initial speed. Figure b shows the trajectories of projectiles launched at the same initial speed of 50 meters per second and different launch angles. The trajectories are shown from launch to landing back at the initial elevation. In orange is the trajectory for an angle of 15 degrees above the horizontal, giving a range R of 128 m. In purple is the trajectory for an angle of 45 degrees above the horizontal, giving a range R of 255 m. In blue is the trajectory for an angle of 75 degrees above the horizontal, giving a range R of 128 m, the same as for the 15 degree trajectory. The maximum height increases with launch angle.

Example 4.9: Comparing golf shots

A golfer finds himself in two different situations on different holes. On the second hole he is 120 m from the green and wants to hit the ball 90 m and let it run onto the green. He angles the shot low to the ground at 30° to the horizontal to let the ball roll after impact. On the fourth hole he is 90 m from the green and wants to let the ball drop with a minimum amount of rolling after impact. Here, he angles the shot at 70° to the horizontal to minimize rolling after impact. Both shots are hit and impacted on a level surface. (a) What is the initial speed of the ball at the second hole? (b) What is the initial speed of the ball at the fourth hole? (c) Write the trajectory equation for both cases. (d) Graph the trajectories.

We see that the range equation has the initial speed and angle, so we can solve for the initial speed for both (a) and (b). When we have the initial speed, we can use this value to write the trajectory equation.

  • $$R = \frac{v_{0}^{2} \sin 2 \theta_{0}}{g} \Rightarrow v_{0} = \sqrt{\dfrac{Rg}{\sin 2\theta_{0}}} = \sqrt{\dfrac{(90.0\; m)(9.8\; m/s^{2})}{\sin (2(30^{o}))}} = 31.9\; m/s$$
  • $$R = \frac{v_{0}^{2} \sin 2 \theta_{0}}{g} \Rightarrow v_{0} = \sqrt{\dfrac{Rg}{\sin 2\theta_{0}}} = \sqrt{\dfrac{(90.0\; m)(9.8\; m/s^{2})}{\sin (2(70^{o}))}} = 37.0\; m/s$$
  • $$y = x \Big[ \tan \theta_{0} - \frac{g}{2(v_{0} \cos \theta_{0})^{2}} x \Big]$$Second hole: $$y = x \Big[ \tan 30^{o} - \frac{9.8\; m/s^{2}}{2[(31.9\; m/s)(\cos 30^{o})]^{2}} x \Big] = 0.58x - 0.0064x^{2}$$Fourth hole: $$y = x \Big[ \tan 70^{o} - \frac{9.8\; m/s^{2}}{2[(37.0\; m/s)(\cos 70^{o})]^{2}} x \Big] = 2.75x - 0.0306x^{2}$$
  • Using a graphing utility, we can compare the two trajectories, which are shown in Figure \(\PageIndex{6}\).

Two parabolic functions are shown. The range for both trajectories is 90 meters. One shot travels much higher than the other. The higher shot has an initial velocity of 37 meters per second and an angle of 70 degrees. The lower shot has an initial velocity of 31.9 meters per second and an angle of 30 degrees.

The initial speed for the shot at 70° is greater than the initial speed of the shot at 30°. Note from Figure \(\PageIndex{6}\) that two projectiles launched at the same speed but at different angles have the same range if the launch angles add to 90°. The launch angles in this example add to give a number greater than 90°. Thus, the shot at 70° has to have a greater launch speed to reach 90 m, otherwise it would land at a shorter distance.

Exercise 4.4

If the two golf shots in Example 4.9 were launched at the same speed, which shot would have the greatest range?

When we speak of the range of a projectile on level ground, we assume R is very small compared with the circumference of Earth. If, however, the range is large, Earth curves away below the projectile and the acceleration resulting from gravity changes direction along the path. The range is larger than predicted by the range equation given earlier because the projectile has farther to fall than it would on level ground, as shown in Figure \(\PageIndex{7}\), which is based on a drawing in Newton’s Principia . If the initial speed is great enough, the projectile goes into orbit. Earth’s surface drops 5 m every 8000 m. In 1 s an object falls 5 m without air resistance. Thus, if an object is given a horizontal velocity of 8000 m/s (or 18,000 mi/hr) near Earth’s surface, it will go into orbit around the planet because the surface continuously falls away from the object. This is roughly the speed of the Space Shuttle in a low Earth orbit when it was operational, or any satellite in a low Earth orbit. These and other aspects of orbital motion, such as Earth’s rotation, are covered in greater depth in Gravitation .

The figure shows a drawing of the earth with a tall tower at the north pole and a horizontal arrow labeled v 0 pointing to the right. 5 trajectories that start at the top of the tower are shown. The first reaches the earth near the tower. The second reaches the earth farther from the tower, and the third even farther. The fourth trajectory hits the earth at the equator, and is tangent to the surface at the equator. The fifth trajectory is a circle concentric with the earth.

At PhET Explorations: Projectile Motion , learn about projectile motion in terms of the launch angle and initial velocity.

Logo for UA Open Textbooks

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

24 Projectile Motion

So far we have only discussed objects in free fall that are going up and down.  However, objects in free fall can also have motion along the horizontal direction. This is often referred to as  projectile motion .  Let’s look at one such example.

Exercise 24.1: Kicking a Ball

v_0

Since this is our first problem in 2D problem involving forces, we will go through it step by step.

Now our picture.  I find it helpful to trace out the path of the ball, which we know goes up and down in an arc.  If you’re not sure how I got my picture, I also added a video in which I go through the whole process.

Now find relations. Since this is a problem involving motion in 2D,  we will need to write Newton’s laws for each object and for each axis!

x

Use the relations we just derived to determine how far the ball travels.

Use your to determine at what angle you should kick the ball to achieve the longest possible kick.  You could do this with calculus, but I think it is much more gratifying if you just reason it out in English.

I want to emphasize two things about our solution above:

  • When solving force problems, we always need to write Newton’s law for each object, for each axis.
  • Once we have our forces set, we can treat the two axis independently of each other .

Let’s do another projectile motion problem.

Exercise 24.2: Cannon Shot

Follow the 4 steps of problem solving.  You can peak at my solution if you get stuck, but then try to finish up the problem on your own.

To end, let’s try a more conceptual question (no math needed!).

Exercise 24.3: Bullet Fired vs Bullet Dropped

A bullet is fired from a gun that is pointed perfectly horizontally.  At the exact same time, a second bullet is dropped from the exact same height.

Fortunately for us, the Myth Busters performed this exact experiment.

This problem highlights a point we raised earlier: the kinematic equations for one axis are completely independent of what is happening in other axis .  The “dropping” motion of the bullet is along the y-axis, so whatever is going on along the x-axis (i.e. the bullet being fired) is completely irrelevant .

Key Takeaways

Introductory Physics: Classical Mechanics Copyright © by . All Rights Reserved.

Share This Book

Physics Problems with Solutions

  • Projectile Problems with Solutions and Explanations

Projectile problems are presented along with detailed solutions . These problems may be better understood when projectile equations are first reviewed. An interactive html 5 applet may be used to better understand the projectile equations.

Problems with Detailed Solutions

An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal. a) What is the maximum height reached by the object? b) What is the total flight time (between launch and touching the ground) of the object? c) What is the horizontal range (maximum x above ground) of the object? d) What is the magnitude of the velocity of the object just before it hits the ground? Solution to Problem 1

A ball kicked from ground level at an initial velocity of 60 m/s and an angle θ with ground reaches a horizontal distance of 200 meters. a) What is the size of angle θ? b) What is time of flight of the ball? Solution to Problem 5

A ball of 600 grams is kicked at an angle of 35° with the ground with an initial velocity V 0 . a) What is the initial velocity V 0 of the ball if its kinetic energy is 22 Joules when its height is maximum? b) What is the maximum height reached by the ball Solution to Problem 6

A projectile starting from ground hits a target on the ground located at a distance of 1000 meters after 40 seconds. a) What is the size of the angle θ? b) At what initial velocity was the projectile launched? Solution to Problem 7

The trajectory of a projectile launched from ground is given by the equation y = -0.025 x 2 + 0.5 x, where x and y are the coordinate of the projectile on a rectangular system of axes. a) Find the initial velocity and the angle at which the projectile is launched. Solution to Problem 8

Two balls A and B of masses 100 grams and 300 grams respectively are pushed horizontally from a table of height 3 meters. Ball has is pushed so that its initial velocity is 10 m/s and ball B is pushed so that its initial velocity is 15 m/s. a) Find the time it takes each ball to hit the ground. b) What is the difference in the distance between the points of impact of the two balls on the ground? Solution to Problem 9

More References and Links

  • Projectile Motion Calculator and Solver
  • Solutions and Explanations to Projectile Problems
  • Projectile Equations with Explanations
  • Interactive Simulation of Projectile .

Popular Pages

  • Privacy Policy

Logo for University of Central Florida Pressbooks

Chapter 3 Two-Dimensional Kinematics

3.4 Projectile Motion

  • Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory.
  • Determine the location and velocity of a projectile at different points in its trajectory.
  • Apply the principle of independence of motion to solve projectile motion problems.

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile , and its path is called its trajectory . The motion of falling objects, as covered in Chapter 2.6 Problem-Solving Basics for One-Dimensional Kinematics , is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible .

\textbf{s}

REVIEW OF KINEMATIC EQUATIONS (CONSTANT α )

\boldsymbol{x=x_0+\bar{v}t}

Given these assumptions, the following steps are then used to analyze projectile motion:

\boldsymbol{\textbf{A}_x=\textbf{A cos}\:\theta}

Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:

\boldsymbol{\textbf{Horizontal Motion}(a_x=0)}

Total displacement and velocity

\boldsymbol{s=\sqrt{x^2+y^2}}

Example 1: A Fireworks Projectile Explodes High and Away

\boldsymbol{75.0^o}

Solution for (a)

\boldsymbol{y}

Discussion for (a)

Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height.

Solution for (b)

\boldsymbol{y=y_0+\frac{1}{2}(v_{0y}+v_y)t}.

Discussion for (b)

\boldsymbol{y=y_0+v_{0y}t-\frac{1}{2}gt^2},

Solution for (c)

\boldsymbol{x=x_0+v_xt},

Discussion for (c)

The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below.

\boldsymbol{y=h};

This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity.

DEFINING A COORDINATE SYSTEM

\boldsymbol{x_0=0}

Example 2: Calculating Projectile Motion: Hot Rock Projectile

\boldsymbol{35.0^o}

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

\boldsymbol{gt^2.}

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

\boldsymbol{\theta_0}

The final vertical velocity is given by the following equation:

\boldsymbol{v_y=v_{0y}-gt,}

which gives

\boldsymbol{v=31.9\textbf{ m/s}.}

Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Chapter 3.5 Addition of Velocities , we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic.

A figure of the Earth is shown and on top of it a very high tower is placed. A projectile satellite is launched from this very high tower with initial velocity of v zero in the horizontal direction. Several trajectories are shown with increasing range. A circular trajectory is shown indicating the satellite achieved its orbit and it is revolving around the Earth.

PHET EXPLORATIONS: PROJECTILE MOTION

Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass. Add air resistance. Make a game out of this simulation by trying to hit a target.

image

  • Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity.

\boldsymbol{y},

Conceptual Questions

\boldsymbol{0^0}

3: For a fixed initial speed, the range of a projectile is determined by the angle at which it is fired. For all but the maximum, there are two angles that give the same range. Considering factors that might affect the ability of an archer to hit a target, such as wind, explain why the smaller angle (closer to the horizontal) is preferable. When would it be necessary for the archer to use the larger angle? Why does the punter in a football game use the higher trajectory?

4: During a lecture demonstration, a professor places two coins on the edge of a table. She then flicks one of the coins horizontally off the table, simultaneously nudging the other over the edge. Describe the subsequent motion of the two coins, in particular discussing whether they hit the floor at the same time.

Problems & Exercises

\boldsymbol{30.0^0}

2: A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction. (a) At what speed does the ball hit the ground? (b) For how long does the ball remain in the air? (c)What maximum height is attained by the ball?

3: A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

\boldsymbol{32^0}

5: An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

6: A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand. (a) At what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used? (b) What other angle gives the same range, and why would it not be used? (c) How long did this pass take?

\boldsymbol{\theta=45^0}

8: Verify the ranges shown for the projectiles in Figure 5 (b) for an initial velocity of 50 m/s at the given initial angles.

\boldsymbol{6.37\times10^3\textbf{ km}}.

12: The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions.

\boldsymbol{25^0}

15: Gun sights are adjusted to aim high to compensate for the effect of gravity, effectively making the gun accurate only for a specific range. (a) If a gun is sighted to hit targets that are at the same height as the gun and 100.0 m away, how low will the bullet hit if aimed directly at a target 150.0 m away? The muzzle velocity of the bullet is 275 m/s. (b) Discuss qualitatively how a larger muzzle velocity would affect this problem and what would be the effect of air resistance.

16: An eagle is flying horizontally at a speed of 3.00 m/s when the fish in her talons wiggles loose and falls into the lake 5.00 m below. Calculate the velocity of the fish relative to the water when it hits the water.

\boldsymbol{30.0^o}

19: Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s.

20: The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 7.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved in solving projectile motion problems.

\boldsymbol{38.0^0}

26: Unreasonable Results (a) Find the maximum range of a super cannon that has a muzzle velocity of 4.0 km/s. (b) What is unreasonable about the range you found? (c) Is the premise unreasonable or is the available equation inapplicable? Explain your answer. (d) If such a muzzle velocity could be obtained, discuss the effects of air resistance, thinning air with altitude, and the curvature of the Earth on the range of the super cannon.

27: Construct Your Own Problem Consider a ball tossed over a fence. Construct a problem in which you calculate the ball’s needed initial velocity to just clear the fence. Among the things to determine are; the height of the fence, the distance to the fence from the point of release of the ball, and the height at which the ball is released. You should also consider whether it is possible to choose the initial speed for the ball and just calculate the angle at which it is thrown. Also examine the possibility of multiple solutions given the distances and heights you have chosen.

\boldsymbol{x=1.30\textbf{ m}\times10^2}

(b) 28.6 m/s (c) 34.3 m/s

\boldsymbol{50.2^0}

(b) The arrow will go over the branch.

\boldsymbol{\textbf{R}=\frac{v_0^2}{\textbf{sin }2\theta_0g}}

(a) 560 m/s

\boldsymbol{8.00 \times 10^3\textbf{ m}}

(c) 80.0 m. This error is not significant because it is only 1% of the answer in part (b).

\boldsymbol{\theta=6.1^0}

yes, the ball lands at 5.3 m from the net

(a) −0.486 m

(b) The larger the muzzle velocity, the smaller the deviation in the vertical direction, because the time of flight would be smaller. Air resistance would have the effect of decreasing the time of flight, therefore increasing the vertical deviation.

4.23 m. No, the owl is not lucky; he misses the nest.

No, the maximum range (neglecting air resistance) is about 92 m.

(a) 24.2 m/s

(b) The ball travels a total of 57.4 m with the brief gust of wind.

\boldsymbol{y-y_0=0=v_{0y}t-\frac{1}{2}gt^2=(v_0\:\textbf{sin}\:\theta)t-\frac{1}{2}gt^2},

College Physics Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

Share This Book

  • 3.4 Projectile Motion
  • Introduction to Science and the Realm of Physics, Physical Quantities, and Units
  • 1.1 Physics: An Introduction
  • 1.2 Physical Quantities and Units
  • 1.3 Accuracy, Precision, and Significant Figures
  • 1.4 Approximation
  • Section Summary
  • Conceptual Questions
  • Problems & Exercises
  • Introduction to One-Dimensional Kinematics
  • 2.1 Displacement
  • 2.2 Vectors, Scalars, and Coordinate Systems
  • 2.3 Time, Velocity, and Speed
  • 2.4 Acceleration
  • 2.5 Motion Equations for Constant Acceleration in One Dimension
  • 2.6 Problem-Solving Basics for One-Dimensional Kinematics
  • 2.7 Falling Objects
  • 2.8 Graphical Analysis of One-Dimensional Motion
  • Introduction to Two-Dimensional Kinematics
  • 3.1 Kinematics in Two Dimensions: An Introduction
  • 3.2 Vector Addition and Subtraction: Graphical Methods
  • 3.3 Vector Addition and Subtraction: Analytical Methods
  • 3.5 Addition of Velocities
  • Introduction to Dynamics: Newton’s Laws of Motion
  • 4.1 Development of Force Concept
  • 4.2 Newton’s First Law of Motion: Inertia
  • 4.3 Newton’s Second Law of Motion: Concept of a System
  • 4.4 Newton’s Third Law of Motion: Symmetry in Forces
  • 4.5 Normal, Tension, and Other Examples of Forces
  • 4.6 Problem-Solving Strategies
  • 4.7 Further Applications of Newton’s Laws of Motion
  • 4.8 Extended Topic: The Four Basic Forces—An Introduction
  • Introduction: Further Applications of Newton’s Laws
  • 5.1 Friction
  • 5.2 Drag Forces
  • 5.3 Elasticity: Stress and Strain
  • Introduction to Uniform Circular Motion and Gravitation
  • 6.1 Rotation Angle and Angular Velocity
  • 6.2 Centripetal Acceleration
  • 6.3 Centripetal Force
  • 6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force
  • 6.5 Newton’s Universal Law of Gravitation
  • 6.6 Satellites and Kepler’s Laws: An Argument for Simplicity
  • Introduction to Work, Energy, and Energy Resources
  • 7.1 Work: The Scientific Definition
  • 7.2 Kinetic Energy and the Work-Energy Theorem
  • 7.3 Gravitational Potential Energy
  • 7.4 Conservative Forces and Potential Energy
  • 7.5 Nonconservative Forces
  • 7.6 Conservation of Energy
  • 7.8 Work, Energy, and Power in Humans
  • 7.9 World Energy Use
  • Introduction to Linear Momentum and Collisions
  • 8.1 Linear Momentum and Force
  • 8.2 Impulse
  • 8.3 Conservation of Momentum
  • 8.4 Elastic Collisions in One Dimension
  • 8.5 Inelastic Collisions in One Dimension
  • 8.6 Collisions of Point Masses in Two Dimensions
  • 8.7 Introduction to Rocket Propulsion
  • Introduction to Statics and Torque
  • 9.1 The First Condition for Equilibrium
  • 9.2 The Second Condition for Equilibrium
  • 9.3 Stability
  • 9.4 Applications of Statics, Including Problem-Solving Strategies
  • 9.5 Simple Machines
  • 9.6 Forces and Torques in Muscles and Joints
  • Introduction to Rotational Motion and Angular Momentum
  • 10.1 Angular Acceleration
  • 10.2 Kinematics of Rotational Motion
  • 10.3 Dynamics of Rotational Motion: Rotational Inertia
  • 10.4 Rotational Kinetic Energy: Work and Energy Revisited
  • 10.5 Angular Momentum and Its Conservation
  • 10.6 Collisions of Extended Bodies in Two Dimensions
  • 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum
  • Introduction to Fluid Statics
  • 11.1 What Is a Fluid?
  • 11.2 Density
  • 11.3 Pressure
  • 11.4 Variation of Pressure with Depth in a Fluid
  • 11.5 Pascal’s Principle
  • 11.6 Gauge Pressure, Absolute Pressure, and Pressure Measurement
  • 11.7 Archimedes’ Principle
  • 11.8 Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action
  • 11.9 Pressures in the Body
  • Introduction to Fluid Dynamics and Its Biological and Medical Applications
  • 12.1 Flow Rate and Its Relation to Velocity
  • 12.2 Bernoulli’s Equation
  • 12.3 The Most General Applications of Bernoulli’s Equation
  • 12.4 Viscosity and Laminar Flow; Poiseuille’s Law
  • 12.5 The Onset of Turbulence
  • 12.6 Motion of an Object in a Viscous Fluid
  • 12.7 Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes
  • Introduction to Temperature, Kinetic Theory, and the Gas Laws
  • 13.1 Temperature
  • 13.2 Thermal Expansion of Solids and Liquids
  • 13.3 The Ideal Gas Law
  • 13.4 Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature
  • 13.5 Phase Changes
  • 13.6 Humidity, Evaporation, and Boiling
  • Introduction to Heat and Heat Transfer Methods
  • 14.2 Temperature Change and Heat Capacity
  • 14.3 Phase Change and Latent Heat
  • 14.4 Heat Transfer Methods
  • 14.5 Conduction
  • 14.6 Convection
  • 14.7 Radiation
  • Introduction to Thermodynamics
  • 15.1 The First Law of Thermodynamics
  • 15.2 The First Law of Thermodynamics and Some Simple Processes
  • 15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency
  • 15.4 Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated
  • 15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators
  • 15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy
  • 15.7 Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation
  • Introduction to Oscillatory Motion and Waves
  • 16.1 Hooke’s Law: Stress and Strain Revisited
  • 16.2 Period and Frequency in Oscillations
  • 16.3 Simple Harmonic Motion: A Special Periodic Motion
  • 16.4 The Simple Pendulum
  • 16.5 Energy and the Simple Harmonic Oscillator
  • 16.6 Uniform Circular Motion and Simple Harmonic Motion
  • 16.7 Damped Harmonic Motion
  • 16.8 Forced Oscillations and Resonance
  • 16.10 Superposition and Interference
  • 16.11 Energy in Waves: Intensity
  • Introduction to the Physics of Hearing
  • 17.2 Speed of Sound, Frequency, and Wavelength
  • 17.3 Sound Intensity and Sound Level
  • 17.4 Doppler Effect and Sonic Booms
  • 17.5 Sound Interference and Resonance: Standing Waves in Air Columns
  • 17.6 Hearing
  • 17.7 Ultrasound
  • Introduction to Electric Charge and Electric Field
  • 18.1 Static Electricity and Charge: Conservation of Charge
  • 18.2 Conductors and Insulators
  • 18.3 Coulomb’s Law
  • 18.4 Electric Field: Concept of a Field Revisited
  • 18.5 Electric Field Lines: Multiple Charges
  • 18.6 Electric Forces in Biology
  • 18.7 Conductors and Electric Fields in Static Equilibrium
  • 18.8 Applications of Electrostatics
  • Introduction to Electric Potential and Electric Energy
  • 19.1 Electric Potential Energy: Potential Difference
  • 19.2 Electric Potential in a Uniform Electric Field
  • 19.3 Electrical Potential Due to a Point Charge
  • 19.4 Equipotential Lines
  • 19.5 Capacitors and Dielectrics
  • 19.6 Capacitors in Series and Parallel
  • 19.7 Energy Stored in Capacitors
  • Introduction to Electric Current, Resistance, and Ohm's Law
  • 20.1 Current
  • 20.2 Ohm’s Law: Resistance and Simple Circuits
  • 20.3 Resistance and Resistivity
  • 20.4 Electric Power and Energy
  • 20.5 Alternating Current versus Direct Current
  • 20.6 Electric Hazards and the Human Body
  • 20.7 Nerve Conduction–Electrocardiograms
  • Introduction to Circuits and DC Instruments
  • 21.1 Resistors in Series and Parallel
  • 21.2 Electromotive Force: Terminal Voltage
  • 21.3 Kirchhoff’s Rules
  • 21.4 DC Voltmeters and Ammeters
  • 21.5 Null Measurements
  • 21.6 DC Circuits Containing Resistors and Capacitors
  • Introduction to Magnetism
  • 22.1 Magnets
  • 22.2 Ferromagnets and Electromagnets
  • 22.3 Magnetic Fields and Magnetic Field Lines
  • 22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field
  • 22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications
  • 22.6 The Hall Effect
  • 22.7 Magnetic Force on a Current-Carrying Conductor
  • 22.8 Torque on a Current Loop: Motors and Meters
  • 22.9 Magnetic Fields Produced by Currents: Ampere’s Law
  • 22.10 Magnetic Force between Two Parallel Conductors
  • 22.11 More Applications of Magnetism
  • Introduction to Electromagnetic Induction, AC Circuits and Electrical Technologies
  • 23.1 Induced Emf and Magnetic Flux
  • 23.2 Faraday’s Law of Induction: Lenz’s Law
  • 23.3 Motional Emf
  • 23.4 Eddy Currents and Magnetic Damping
  • 23.5 Electric Generators
  • 23.6 Back Emf
  • 23.7 Transformers
  • 23.8 Electrical Safety: Systems and Devices
  • 23.9 Inductance
  • 23.10 RL Circuits
  • 23.11 Reactance, Inductive and Capacitive
  • 23.12 RLC Series AC Circuits
  • Introduction to Electromagnetic Waves
  • 24.1 Maxwell’s Equations: Electromagnetic Waves Predicted and Observed
  • 24.2 Production of Electromagnetic Waves
  • 24.3 The Electromagnetic Spectrum
  • 24.4 Energy in Electromagnetic Waves
  • Introduction to Geometric Optics
  • 25.1 The Ray Aspect of Light
  • 25.2 The Law of Reflection
  • 25.3 The Law of Refraction
  • 25.4 Total Internal Reflection
  • 25.5 Dispersion: The Rainbow and Prisms
  • 25.6 Image Formation by Lenses
  • 25.7 Image Formation by Mirrors
  • Introduction to Vision and Optical Instruments
  • 26.1 Physics of the Eye
  • 26.2 Vision Correction
  • 26.3 Color and Color Vision
  • 26.4 Microscopes
  • 26.5 Telescopes
  • 26.6 Aberrations
  • Introduction to Wave Optics
  • 27.1 The Wave Aspect of Light: Interference
  • 27.2 Huygens's Principle: Diffraction
  • 27.3 Young’s Double Slit Experiment
  • 27.4 Multiple Slit Diffraction
  • 27.5 Single Slit Diffraction
  • 27.6 Limits of Resolution: The Rayleigh Criterion
  • 27.7 Thin Film Interference
  • 27.8 Polarization
  • 27.9 *Extended Topic* Microscopy Enhanced by the Wave Characteristics of Light
  • Introduction to Special Relativity
  • 28.1 Einstein’s Postulates
  • 28.2 Simultaneity And Time Dilation
  • 28.3 Length Contraction
  • 28.4 Relativistic Addition of Velocities
  • 28.5 Relativistic Momentum
  • 28.6 Relativistic Energy
  • Introduction to Quantum Physics
  • 29.1 Quantization of Energy
  • 29.2 The Photoelectric Effect
  • 29.3 Photon Energies and the Electromagnetic Spectrum
  • 29.4 Photon Momentum
  • 29.5 The Particle-Wave Duality
  • 29.6 The Wave Nature of Matter
  • 29.7 Probability: The Heisenberg Uncertainty Principle
  • 29.8 The Particle-Wave Duality Reviewed
  • Introduction to Atomic Physics
  • 30.1 Discovery of the Atom
  • 30.2 Discovery of the Parts of the Atom: Electrons and Nuclei
  • 30.3 Bohr’s Theory of the Hydrogen Atom
  • 30.4 X Rays: Atomic Origins and Applications
  • 30.5 Applications of Atomic Excitations and De-Excitations
  • 30.6 The Wave Nature of Matter Causes Quantization
  • 30.7 Patterns in Spectra Reveal More Quantization
  • 30.8 Quantum Numbers and Rules
  • 30.9 The Pauli Exclusion Principle
  • Introduction to Radioactivity and Nuclear Physics
  • 31.1 Nuclear Radioactivity
  • 31.2 Radiation Detection and Detectors
  • 31.3 Substructure of the Nucleus
  • 31.4 Nuclear Decay and Conservation Laws
  • 31.5 Half-Life and Activity
  • 31.6 Binding Energy
  • 31.7 Tunneling
  • Introduction to Applications of Nuclear Physics
  • 32.1 Diagnostics and Medical Imaging
  • 32.2 Biological Effects of Ionizing Radiation
  • 32.3 Therapeutic Uses of Ionizing Radiation
  • 32.4 Food Irradiation
  • 32.5 Fusion
  • 32.6 Fission
  • 32.7 Nuclear Weapons
  • Introduction to Particle Physics
  • 33.1 The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited
  • 33.2 The Four Basic Forces
  • 33.3 Accelerators Create Matter from Energy
  • 33.4 Particles, Patterns, and Conservation Laws
  • 33.5 Quarks: Is That All There Is?
  • 33.6 GUTs: The Unification of Forces
  • Introduction to Frontiers of Physics
  • 34.1 Cosmology and Particle Physics
  • 34.2 General Relativity and Quantum Gravity
  • 34.3 Superstrings
  • 34.4 Dark Matter and Closure
  • 34.5 Complexity and Chaos
  • 34.6 High-temperature Superconductors
  • 34.7 Some Questions We Know to Ask
  • A | Atomic Masses
  • B | Selected Radioactive Isotopes
  • C | Useful Information
  • D | Glossary of Key Symbols and Notation

Learning Objectives

By the end of this section, you will be able to:

  • Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory.
  • Determine the location and velocity of a projectile at different points in its trajectory.
  • Apply the principle of independence of motion to solve projectile motion problems.

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile , and its path is called its trajectory . The motion of falling objects, as covered in Problem-Solving Basics for One-Dimensional Kinematics , is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible .

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction , where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x -axis and the vertical axis the y -axis. Figure 3.34 illustrates the notation for displacement, where s s is defined to be the total displacement and x x and y y are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s , x , and y . (Note that in the last section we used the notation A A to represent a vector with components A x A x and A y A y . If we continued this format, we would call displacement s s with components s x s x and s y s y . However, to simplify the notation, we will simply represent the component vectors as x x and y y .)

Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x - and y -axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: a y = – g = – 9.80 m /s 2 a y = – g = – 9.80 m /s 2 . (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, a x = 0 a x = 0 . Both accelerations are constant, so the kinematic equations can be used.

Review of Kinematic Equations (constant a a )

Given these assumptions, the following steps are then used to analyze projectile motion:

Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so A x = A cos θ A x = A cos θ and A y = A sin θ A y = A sin θ are used. The magnitude of the components of displacement s s along these axes are x x and y. y. The magnitudes of the components of the velocity v v are v x = v cos θ v x = v cos θ and v y = v sin θ, v y = v sin θ, where v v is the magnitude of the velocity and θ θ is its direction, as shown in Figure 3.35 . Initial values are denoted with a subscript 0, as usual.

Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:

Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time t t . The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below.

Step 4. Recombine the two motions to find the total displacement s s and velocity v v . Because the x - and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing A = A x 2 + A y 2 A = A x 2 + A y 2 and θ = tan − 1 ( A y / A x ) θ = tan − 1 ( A y / A x ) in the following form, where θ θ is the direction of the displacement s s and θ v θ v is the direction of the velocity v v :

Total displacement and velocity

Example 3.4

A fireworks projectile explodes high and away.

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0º 75.0º above the horizontal, as illustrated in Figure 3.36 . The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which a x = 0 a x = 0 and a y = – g a y = – g . We can then define x 0 x 0 and y 0 y 0 to be zero and solve for the desired quantities.

Solution for (a)

By “height” we mean the altitude or vertical position y y above the starting point. The highest point in any trajectory, called the apex, is reached when v y = 0 v y = 0 . Since we know the initial and final velocities as well as the initial position, we use the following equation to find y y :

Because y 0 y 0 and v y v y are both zero, the equation simplifies to

Solving for y y gives

Now we must find v 0 y v 0 y , the component of the initial velocity in the y -direction. It is given by v 0 y = v 0 sin θ v 0 y = v 0 sin θ , where v 0 y v 0 y is the initial velocity of 70.0 m/s, and θ 0 = 75.0º θ 0 = 75.0º is the initial angle. Thus,

Discussion for (a)

Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height.

Solution for (b)

As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use y = y 0 + 1 2 ( v 0 y + v y ) t y = y 0 + 1 2 ( v 0 y + v y ) t . Because y 0 y 0 is zero, this equation reduces to simply

Note that the final vertical velocity, v y v y , at the highest point is zero. Thus,

Discussion for (b)

This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. (Another way of finding the time is by using y = y 0 + v 0 y t − 1 2 gt 2 y = y 0 + v 0 y t − 1 2 gt 2 , and solving the quadratic equation for t t .)

Solution for (c)

Because air resistance is negligible, a x = 0 a x = 0 and the horizontal velocity is constant, as discussed above. The horizontal displacement is horizontal velocity multiplied by time as given by x = x 0 + v x t x = x 0 + v x t , where x 0 x 0 is equal to zero:

where v x v x is the x -component of the velocity, which is given by v x = v 0 cos θ 0 . v x = v 0 cos θ 0 . Now,

The time t t for both motions is the same, and so x x is

Discussion for (c)

The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below.

In solving part (a) of the preceding example, the expression we found for y y is valid for any projectile motion where air resistance is negligible. Call the maximum height y = h y = h ; then,

This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity.

Defining a Coordinate System

It is important to set up a coordinate system when analyzing projectile motion. One part of defining the coordinate system is to define an origin for the x x and y y positions. Often, it is convenient to choose the initial position of the object as the origin such that x 0 = 0 x 0 = 0 and y 0 = 0 y 0 = 0 . It is also important to define the positive and negative directions in the x x and y y directions. Typically, we define the positive vertical direction as upwards, and the positive horizontal direction is usually the direction of the object’s motion. When this is the case, the vertical acceleration, a y = – g a y = – g , takes a negative value (since it is directed downwards towards the Earth). However, it is occasionally useful to define the coordinates differently. For example, if you are analyzing the motion of a ball thrown downwards from the top of a cliff, it may make sense to define the positive direction downwards since the motion of the ball is solely in the downwards direction. If this is the case, a y = g a y = g takes a positive value.

Example 3.5

Calculating projectile motion: hot rock projectile.

Kilauea in Hawaii is the world’s most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 25.0 m/s and at an angle 35.0º 35.0º above the horizontal, as shown in Figure 3.37 . The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and direction of the rock’s velocity at impact?

Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. We will solve for t t first. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be recombined to obtain v v and θ v θ v at the final time t t determined in the first part of the example.

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

If we take the initial position y 0 y 0 to be zero, then the final position is y = − 20 .0 m . y = − 20 .0 m . Now the initial vertical velocity is the vertical component of the initial velocity, found from v 0 y = v 0 sin θ 0 v 0 y = v 0 sin θ 0 = ( 25 . 0 m/s 25 . 0 m/s )( sin 35.0º sin 35.0º ) = 14 . 3 m/s 14 . 3 m/s . Substituting known values yields

Rearranging terms gives a quadratic equation in t t :

This expression is a quadratic equation of the form at 2 + bt + c = 0 at 2 + bt + c = 0 , where the constants are a = 4.90 a = 4.90 , b = – 14.3 b = – 14.3 , and c = – 20.0. c = – 20.0. Its solutions are given by the quadratic formula:

This equation yields two solutions: t = 3.96 t = 3.96 and t = – 1.03 t = – 1.03 . (It is left as an exercise for the reader to verify these solutions.) The time is t = 3.96 s t = 3.96 s or – 1.03 s – 1.03 s . The negative value of time implies an event before the start of motion, and so we discard it. Thus,

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

From the information now in hand, we can find the final horizontal and vertical velocities v x v x and v y v y and combine them to find the total velocity v v and the angle θ 0 θ 0 it makes with the horizontal. Of course, v x v x is constant so we can solve for it at any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle. Therefore:

The final vertical velocity is given by the following equation:

where v 0y v 0y was found in part (a) to be 14 . 3 m/s 14 . 3 m/s . Thus,

To find the magnitude of the final velocity v v we combine its perpendicular components, using the following equation:

which gives

The direction θ v θ v is found from the equation:

The negative angle means that the velocity is 50 . 1º 50 . 1º below the horizontal. This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See Figure 3.37 .)

One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each other. Galileo was the first person to fully comprehend this characteristic. He used it to predict the range of a projectile. On level ground, we define range to be the horizontal distance R R traveled by a projectile. Galileo and many others were interested in the range of projectiles primarily for military purposes—such as aiming cannons. However, investigating the range of projectiles can shed light on other interesting phenomena, such as the orbits of satellites around the Earth. Let us consider projectile range further.

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed v 0 v 0 , the greater the range, as shown in Figure 3.38 (a). The initial angle θ 0 θ 0 also has a dramatic effect on the range, as illustrated in Figure 3.38 (b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with θ 0 = 45º θ 0 = 45º . This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately 38º 38º . Interestingly, for every initial angle except 45º 45º , there are two angles that give the same range—the sum of those angles is 90º 90º . The range also depends on the value of the acceleration of gravity g g . The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range R R of a projectile on level ground for which air resistance is negligible is given by

where v 0 v 0 is the initial speed and θ 0 θ 0 is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described.

When we speak of the range of a projectile on level ground, we assume that R R is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See Figure 3.39 .) If the initial speed is great enough, the projectile goes into orbit. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text.

Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Addition of Velocities , we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic.

PhET Explorations

Projectile motion.

Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass. Add air resistance. Make a game out of this simulation by trying to hit a target.

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units
  • Authors: Paul Peter Urone, Roger Hinrichs
  • Publisher/website: OpenStax
  • Book title: College Physics 2e
  • Publication date: Jul 13, 2022
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units
  • Section URL: https://openstax.org/books/college-physics-2e/pages/3-4-projectile-motion

© Jan 19, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

Physexams Logo

  • Exam Center
  • Ticket Center
  • Flash Cards
  • Straight Line Motion

Grade 12: Physics Worksheet on Projectile Motion

Looking to master projectile motion in your physics class? Check out our comprehensive worksheet with detailed solutions to help you understand the concepts and excel in your studies!

There is also a pdf  for worksheets with answers.

A summary of projectile motion:

Any motion having the following conditions is called a projectile motion.     (i) Follows a parabolic path (trajectory).     (ii) Moves under the influence of gravity. 

The projectile motion formulas applied to solve two-dimensional projectile motion problems are as follows \begin{gather*} x=(v_0\cos\theta)t+x_0\\\\y=-\frac 12 gt^2+(v_0\sin\theta)t+y_0\\\\ v_y=v_0\sin\theta-gt\\\\v_y^2-(v_0\sin\theta)^2=-2g(y-y_0)\end{gather*} 

Projectile motion problems and answers

Problem (1): A person kicks a ball with an initial velocity of 15 m/s at an angle of 37° above the horizontal (neglecting the air resistance). Find  (a) the total time the ball is in the air.  (b) the horizontal distance traveled by the ball

Solution : The initial step in answering any projectile motion questions is to establish a coordinate system and sketch the path of the projectile, including its initial and final positions and velocities. 

By doing so, you will be able to solve the relevant projectile equations easily. 

Hence, we choose the origin of the coordinate system to be at the throwing point, $x_0=0, y_0=0$. 

Sketch of a projectile motion

(a) Here, we are seeking the time it takes for the projectile to travel from the point of release to the ground.  

In reality, projectiles have two independent motions: one in the horizontal direction with uniform motion at a constant velocity (i.e., $a_x=0$), and the other in the vertical direction under the effect of gravity (with $a_y=-g$).  

The kinematic equations that describe the horizontal and vertical distances are as follows:  \begin{gather*} x=x_0+(\underbrace{v_0\cos \theta}_{v_{0x}})t \\ y=-\frac 12 gt^2+(\underbrace{v_0\sin \theta}_{v_{0y}})t+y_0\end{gather*} By substituting the coordinates of the initial and final points into the vertical equation, we can find the total time the ball is in the air.

Setting $y=0$ in the second equation (because the projectile lands at the same level as the throwing point.), we have: \begin{align*} y&=-\frac 12 gt^2+(v_0\sin \theta)t+y_0\\\\ 0&=-\frac 12 (9.8)t^2+(15)\sin 37^\circ\,t+0 \end{align*} By rearranging the above expression, we can obtain two solutions for $t$: \begin{gather*} t_1=0 \\\\ t_2=\frac{2\times 15\sin37^\circ}{9.8}=1.84\,{\rm s}\end{gather*} The first time represents the starting moment, and the second represents the total time the ball was in the air. 

(b) As previously mentioned, projectile motion consists of two independent motions with different positions, velocities, and accelerations, each described by distinct kinematic equations. 

The time it takes for the projectile to reach a specific point horizontally is equal to the time it takes to fall vertically to that point. Therefore, time is the common factor in both the horizontal and vertical motions of a projectile.

In this particular problem, the time calculated in part (a) can be used in the horizontal kinematic equation to determine the distance traveled. 

Substituting the obtained time into the equation, we find that the projectile travels a distance of 22.08 meters. \begin{align*} x&=x_0+(v_0\cos \theta)t \\ &=0+(15)\cos 37^\circ\,(1.84) \\ &=22.08\quad {\rm m}\end{align*}

Be sure to check the following questions.

If you are getting ready for the AP Physics exam, these problems on the kinematics equation (or these AP Physics Kinematics Problems ) are probably also useful to you. 

Problem (2): A ball is thrown into the air at an angle of 60° above the horizontal with an initial velocity of 40 m/s from a  50-m-high building. Find  (a) The time to reach the top of its path. (b) The maximum height the ball reached from the base of the building.

Solution : In the previous question, we found that the motion of a projectile consists of two distinct vertical and horizontal movements.

We learned about how to find distances in both directions using relevant kinematic equations. 

There is also another set of kinematic equations that discuss the velocities in vertical and horizontal directions as follows \begin{gather*} v_x=v_{0x}=v_0\cos\theta \\ v_y=v_{0y}-gt=v_0\sin\theta-gt\end{gather*} As you can see, the horizontal component of the velocity, $v_x$, is constant throughout the motion, but the vertical component varies with time. 

As an important note, keep in mind that in the problems about projectile motions, at the highest point of the trajectory, the vertical component of the velocity is always zero, i.e., $v_y=0$. 

To solve the first part of the problem, specify two points as the initial and final points, then solve the relevant kinematic equations between those points.

Here, setting $v_y=0$ in the second equation and solving for the unknown time $t$, we have \begin{align*} v_y&=v_{0y}-gt \\\\ &=v_0\sin\theta-gt \\\\ 0&=(40)(\sin 60^\circ)-(9.8)(t) \\\\ \Rightarrow t&=\boxed{3.53\, {\rm s}}\end{align*} Thus, the time taken to reach the maximum height of the trajectory (path) is 3.53 seconds from the moment of the launch. We call this maximum time $t_{max}$. 

A projectile motion problems from a building

(b) Let the origin be at the throwing point, so $x_0=y_0=0$ in the kinematic equations. In this part, the vertical distance traveled to the maximum point is requested. 

By substituting $t_{max}$ into the vertical distance projectile equation, we can find the maximum height as below \begin{align*} y-y_0&=-\frac 12 gt^2 +(v_0 \sin \theta)t\\ \\ y_{max}&=-\frac 12 (9.8)(3.53)^2+(40 \sin 60^\circ)(3.53)\\\\ &=61.22\quad {\rm m}\end{align*} Therefore, the maximum height at which the ball is reached from the base of the building is \[H=50+y_{max}=111.22\quad {\rm m}\] 

For further reading about uniform motion along the horizontal direction, see  speed, velocity, and acceleration problems .

Problem (3): A person standing on the edge of a 50-m-high cliff throws a stone horizontally at a speed of 18 m/s.  (a) What is the initial position of the stone?  (b) What are the components of the initial velocity? (c) What are the $x$- and $y$-components of the velocity of the stone at any arbitrary time $t$? (d) How long will it take the stone to strike the bottom of the cliff? (e) With what angle and speed does the stone strike the ground below the cliff?

projectile motion problem from a high cliff

Solution : As mentioned repeatedly, as a first step to solving a projectile motion problem, choose a relevant coordinate system. 

(a) Usually, we place the origin of the coordinate system at the point where the projectile is thrown. In this case, the coordinate of the initial position is $x_0=0\, , \, y_0=0$.

If we had chosen the coordinate at the base of the cliff and placed the origin at that point, the position of the initial point would have been $x_0=0\, , \, y_0=50\,\rm m$. 

projectile motion problem - Solution with an arbitrary coordinate system

(b) The stone is thrown horizontally with $\theta=0$, so there is no vertical velocity component. Consequently, its initial speed components are $v_{0x}=18\,\rm m/s$ and $v_{0y}=0$. 

(c) Remember that the velocity components of a projectile change over time according to the following formula. Substituting the given values into it, we get:  \begin{align*} v_x&=v_{0x}=18\,\rm m/s \\\\ v_y&=v_{0y}-gt\\&=-9.8t \end{align*}  (d) If we take the top of the cliff to be the origin of our coordinate system, with  $x_0=y_0=0$, the stone hits the ground $50\,\rm m$ below our chosen origin. Therefore, the coordinates of that point would be $x_0=?\, ,\, y=-50\,\rm m$. 

Use the vertical displacement kinematic equation below, substitute the numerical values into that, and solve for time $t$ get \begin{gather*} y-y_0=-\frac 12 gt^2+v_{0y}t \\\\ -50-0=-\frac 12 (9.8)t^2-0 \\\\ \Rightarrow \quad \boxed{t=3.2\,\rm s} \end{gather*}  (e) First, find the velocity components just before the stone hits the ground. We know that it takes about $3.2\,\rm s$ for the stone to reach the bottom of the cliff. Thus, put this time value into the formulas of velocity components at any time $t$. \begin{align*} v_x&=v_{0x}=18\,\rm m/s \\\\ v_y&=v_{0y}-gt\\&=-9.8\times 3.2 \\&=-31.36\,\rm m/s \end{align*} Notice that the negative sign here indicates that the vertical velocity component just before hitting the ground points downward, as expected. 

Recall from the section on  vector practice problems that when we have the components of a vector, we can find its magnitude using the Pythagorean theorem. This gives us: \begin{align*} v&=\sqrt{v_x^2+v_y^2} \\\\ &=\sqrt{18^2+(-31.36)^2} \\ &=\boxed{36.15\,\rm m/s} \end{align*} We can also find the angle of impact with the ground using the velocity components: \begin{align*} \tan\theta&=\frac{v_y}{v_x} \\\\ &=\frac{-31.36}{18} \\\\ &=-1.74 \end{align*} Taking the inverse tangent of both sides gives: \[\theta=\tan^{-1}(-1.74)=\boxed{-60.1^\circ} \] Therefore, the stone hit the ground at an angle of about $60^\circ$ with a speed of $36.15\,\rm m/s$. The negative angle indicates that it is below horizontal, which is to be expected.

Problem (4): A book slides off a frictionless tabletop with a constant speed of 1.1 m/s, and 0.35 seconds later strikes the floor.  (a) How high is the tabletop from the floor? (b) What is the horizontal distance to that point where the book strikes the floor? (c) How fast and at what angle does the book strike the floor? 

Solution : The total time the book is in the air is $t=0.35\,\rm s$, and the initial speed with which the book leaves the table horizontally is $v_{0x}=1.1\,\rm m/s$, and $v_{0y}=0$. 

Contrary to the earlier question, in this case, for practice, we place the origin at the bottom of the tabletop. Thus, the coordinates of the initial position would be $x_0=0\, ,\, y_0=h=?$ and the coordinates of the landing point would be $x=? \, , \, y=0$. 

(a) If we substitute the given values into the vertical displacement kinematic equation, we will have \begin{gather*} y-y_0=-\frac 12 gt^2+v_{0y}t \\\\ 0-h=-\frac 12 (9.8)(1.1)^2+0 \\\\ \Rightarrow \quad \boxed{h=5.93\,\rm m} \end{gather*} Thus, the table is 3-m-high.

(b)  Remember that projectile motion involves two motions that are uniformly accelerated in the horizontal and vertical directions.  The time taken by the projectile in the vertical direction is equal to the horizontal time to the landing point.  

Between the start and final points in the horizontal direction of the projectile path, we can write $\Delta x=v_{0x}t$. Substituting the given values into is get \begin{gather*} x-0=1.1\times (0.35) \\ \Rightarrow \quad \boxed{x=0.385\,\rm m} \end{gather*} where $\Delta x=x-x_0$. Therefore, the book strikes the floor about $39\,\rm cm$ ahead of the base of the table. 

(c) This part is straightforward. Substitute the given numerical values into the velocity vector component formulas as below to find the velocity components just before the book strikes the floor. \begin{align*} v_x&=v_{0x}=1.1\,\rm m/s \\\\ v_y&=v_{0y}-gt \\\\ &=0-(9.8)(0.35) \\&=-3.43\,\rm m/s \end{align*} The square root of the sum of the velocity components squared gives the velocity of the book at the instant of hitting \begin{align*} v&=\sqrt{v_x^2+v_y^2} \\\\ &=\sqrt{1.1^2+(-3.43)^2} \\\\ &=\boxed{3.6\,\rm m/s} \end{align*}  and angle of impact with the ground is found as below \begin{align*} \theta&=\tan^{-1}\left(\frac{v_y}{v_x}\right) \\\\ &=\tan^{-1}\left(\frac{-3.43}{1.1}\right) \\\\ &=\tan^{-1}(-3.11) \\\\&=\boxed{-72.17^\circ} \end{align*} Overall, the book hit the ground at an angle of about $72^\circ$ with a speed of nearly $3.6\,\rm m/s$. 

Problem (5): A cannonball is fired from a cliff with a speed of 800 m/s at an angle of 30° below the horizontal. How long will it take to reach 150 m below the firing point? 

Solution: First, choose the origin to be at the firing point, so $x_0=y_0=0$. Now, list the known values as follows      (i) Projectile's initial velocity = $800\,{\rm m/s}$      (ii) Angle of projectile: $-30^\circ$, the negative sign is because the throw is below the horizontal.     (iii) y-coordinate of the final point, 150 m below the origin, $y=-150\,{\rm m}$.       In this problem, the time it takes for the cannonball to reach 100 m below the starting point is required. 

Since the displacement to that point is known, we apply the vertical displacement projectile formula to find the needed time as below \begin{align*} y-y_0&=-\frac 12 gt^2+(v_0\sin\theta)t\\\\-150&=-\frac 12 (9.8)t^2+(800)\sin(-30^\circ)t\\\\ \Rightarrow & \ 4.9t^2+400t-150=0\end{align*} The above quadratic equation has two solutions, $t_1=0.37\,{\rm s}$ and $t_2=-82\,{\rm s}$. It is obvious that the second time is not acceptable.

Therefore, the cannonball takes 0.37 seconds to reach 150 meters below the firing point. 

Problem (6): Someone throws a stone into the air from ground level. The horizontal component of velocity is 25 m/s and it takes 3 seconds for the stone to come back to the same height as before. Find  (a) The range of the stone. (b) The initial vertical component of velocity (c) The angle of projection. 

Solution: The known values are      (i) The initial horizontal component of velocity, $v_{0x}=25\,{\rm m/s}$.      (ii) The time between the initial and final points, which are at the same level, $t=3\,{\rm s}$. 

(a)  The range of projectile motion is defined as the horizontal distance between the launch point and impact at the same elevation.

Because the horizontal motion in projectiles is a motion with constant velocity, the distance traveled in this direction is obtained as $x=v_{0x}t$, where $v_{0x}$ is the initial component of the velocity. 

If you put the total time the projectile is in the air into this formula, you get the range of the projectile. 

In this problem, the stone is thrown from the ground level and, after 3 seconds, reaches the same height. Thus, this is the total time of the projectile. 

Hence, the range of the stone is found as below \begin{align*} x&=v_{0x}t\\&=(25)(3)\\&=75\,{\rm m}\end{align*}

(b)  The initial vertical component of the projectile's velocity appears in two equations, $v_y=v_{0y}-gt$ and $y-y_0=-\frac 12 gt^2+v_{0y}t$. 

Using the second formula is more straightforward because the stone reaches the same height, so its vertical displacement between the initial and final points is zero, i.e., $y-y_0=0$. Setting this into the vertical distance projectile equation, we get \begin{align*} y-y_0&=-\frac 12 gt^2+v_{0y}t\\\\ 0&=-\frac 12 (9.8)(3)^2+v_{0y}(3) \\\\ \Rightarrow v_{0y}&=14.7\quad{\rm m/s}\end{align*} To use the first formula, we need some extra facts about projectile motion in the absence of air resistance, as below      (i) The vertical velocity is zero at the highest point of the path of the projectile, i.e., $v_y=0$.     (ii) If the projectile lands at the same elevation from which it was launched, then the time it takes to reach the highest point of the trajectory is half the total time between the initial and final points.      The second note, in the absence of air resistance, is only valid. 

In this problem, the total flight time is 3 s because air resistance is negligible, so 1.5 seconds are needed for the stone to reach the maximum height of its path. 

Therefore, using the second equation, we can find $v_{0y}$ as below \begin{align*} v_y&=v_{0y}-gt\\0&=v_{0y}-(9.8)(1.5) \\\Rightarrow v_{0y}&=14.7\quad {\rm m/s}\end{align*}  (c) The projection angle is the angle at which the projectile is thrown into the air and performs a two-dimensional motion. 

Once the components of the initial velocity are available, using trigonometry, we can find the angle of projection as below \begin{align*} \theta&=\tan^{-1}\left(\frac{v_{0y}}{v_{0x}}\right)\\\\&=\tan^{-1}\left(\frac{14.7}{25}\right)\\\\&=+30.45^\circ\end{align*} Therefore, the stone is thrown into the air at an angle of about $30^\circ$ above the horizontal.

Problem (7): A ball is thrown at an angle of 60° with an initial speed of 200 m/s. (Neglect the air resistance.) (a) How long is the ball in the air? (b) Find the maximum horizontal distance traveled by the ball. (c) What is the maximum height reached by the ball?

Solution: We choose the origin to be the ball's initial position so that $x_0=y_0=0$. The given data is      (i) The projection angle: $60^\circ$.     (ii) Initial speed : $v_0=200\,{\rm m/s}$. 

(a) The initial and final points of the ball are at the same level, i.e., $y-y_0=0$. 

Thus, the total time the ball is in the air is found by setting $y=0$ in the projectile equation $y=-\frac 12 gt^2+v_{0y}t$ and solving for time $t$ as below \begin{align*} y&=-\frac 12 gt^2+(v_0\sin\theta)t\\\\0&=-\frac 12 (9.8)t^2+(200)(\sin 60^\circ)t\\\\\Rightarrow \quad & \boxed{(-4.9t+100\sqrt{3})t=0} \end{align*} The above expression has two solutions for $t$. One is the initial time, $t_1=0$, and the other is computed as $t_2=35.4\,{\rm s}$. 

Hence, the ball has been in the air for about 35 s. 

(b) The horizontal distance is called the range of the projectile. By inserting the above time (total flight time) into the horizontal distance projectile equation $x=v_{0x}t$, we can find the desired distance traveled. \begin{align*} x&=(v_0\cos\theta)t\\&=(200)(\cos 60^\circ)(35.4)\\&=3540 \quad {\rm m}\end{align*} Therefore, the ball hits the ground 3540 meters away from the throwing point.

(c) Using the projectile equation $v_y^2-v_{0y}^2=-2g(y-y_0)$, setting $v_y=0$ at the highest point of the path, and solving for the vertical distance $y$, the maximum height is found as follows \begin{align*} v_y^2-v_{0y}^2&=-2g(y-y_0)\\0-(200\sin 60^\circ)^2&=-2(9.8)y\\\Rightarrow y&=1531\quad {\rm m}\end{align*} Another method: As mentioned above, the ball hits the ground at the same level as before, so by knowing the total flight time and halving it, we can find the time it takes the ball to reach the highest point of its trajectory. 

Therefore, by setting the half of the total flight time in the following projectile kinematic formula and solving for $y$, we can find the maximum height as \begin{align*} y-y_0&=-\frac 12 gt^2+(v_0\sin\theta)t\\\\y&=-\frac 12 (9.8)(17.7)^2+(200\sin60^\circ)(17.7)\\\\&=1531\quad {\rm m}\end{align*} Hence, the ball reaches 1531 meters above the launch point. 

Problem (8): What are the horizontal range and maximum height of a bullet fired with a speed of 20 m/s at 30° above the horizontal?

Solution : first, find the total flight time, then insert it into the horizontal displacement projectile equation $x=v_{0x}t$ to find the range of the bullet.

Because the bullet lands at the same level as the original, its vertical displacement is zero, $y-y_0=0$, in the following projectile formula we can find the total flight time \begin{align*} y-y_0&=-\frac 12 gt^2+(v_0\sin\theta)t\\\\0&=-\frac 12 (9.8)t^2+(20)(\sin30^\circ)t\\\\ \Rightarrow & (-4.9t+10)t=0\end{align*} Solving for time $t$ gives two solutions: one is the initial time $t_1=0$, and the other is $t_2=1.02\,{\rm s}$. Thus the total time of flight is 2.04 s. 

Therefore, the maximum horizontal distance traveled by the bullet, which is defined as the range of the projectile, is calculated as \begin{align*} x&=(v_0\cos\theta)t\\&=(20\cos 30^\circ)(2.04)\\&=35.3\quad {\rm m}\end{align*} Hence, the bullet lands about 17 m away from the launch point.

Because the air resistance is negligible and the bullet lands at the same height as the original, the time it takes to reach the highest point of its path is always half the total flight time. 

On the other hand, recall that the vertical component of velocity at the maximum height is always zero, i.e., $v_y=0$. By inserting these two notes into the following projectile equation, we have \begin{align*} y&=-\frac 12 gt^2+(v_0\sin\theta)t\\\\&=-\frac 12 (9.8)(1.02)^2+(20\sin 30^\circ)(1.02)\\\\&=5.1\quad {\rm m}\end{align*} We could also use the kinematic equation $v_y^2-v_{0y}^2=-2g(y-y_0)$, to find the maximum height as below \begin{align*} v_y^2-(v_0 \sin \theta)^2 &=-2g(y-y_0) \\ 0-(20\sin 30^\circ)^2 &=-2(9.8)H\\ \Rightarrow H&= 5.1\quad {\rm m} \end{align*} I think the second method is much simpler. 

Problem (9): A projectile is fired horizontally at a speed of 8 m/s from an 80-m-high cliff. Find  (a) The velocity just before the projectile hits the ground.  (b) The angle of impact.

Solution : In this problem, the angle of the projectile is zero because it is fired horizontally. 

The velocity at each point of a projectile trajectory (path) is obtained by the following formula: \[v=\sqrt{v_x^2+v_y^2}\] where $v_x$ and $v_y$ are the horizontal and vertical components of the projectile's velocity at any instant of time. 

(a) Recall that the horizontal component of the projectile's velocity is always constant and, for this problem, is found as \begin{align*} v_x&=v_0\cos\theta\\&=8\times \cos 0^\circ\\&=8\quad {\rm m/s}\end{align*} To find the vertical component of the projectile velocity at any moment, $v_y=v_0\sin\theta-gt$, we should find the time taken to reach that point. 

In this problem, that point is located just before striking the ground, whose coordinates are $y=-80\,{\rm m}, x=?$.

Because it is below the origin, which is assumed to be at the firing point, we inserted a minus sign. 

Because displacement in the vertical direction is known, we can use the projectile formula for vertical distance. 

By setting $y=-80$ into it and solving for the time $t$ needed the projectile reaches the ground, we get \begin{align*} y&=-\frac 12 gt^2+(v_0\sin\theta)t\\\\-80&=-\frac 12 (9.8)t^2+(8\times \sin 0^\circ)t\\\\\Rightarrow t&=\sqrt{\frac{2(80)}{9.8}}\\\\&=2.86\quad {\rm s}\end{align*} Now insert this time into the $y$-component of the projectile's velocity to find $v_y$ just before hitting the ground \begin{align*} v_y&=v_0\sin\theta-gt\\ &=8\sin 0^\circ-(9.8)(2.86)\\&=-28\quad {\rm m/s}\end{align*} Now that both components of the velocity are available, we can compute its magnitude as below \begin{align*} v&=\sqrt{v_x^2+v_y^2}\\\\&=\sqrt{8^2+(-28)^2}\\\\&=29.1\quad {\rm m/s}\end{align*} Therefore, the projectile hits the ground at a speed of 29.1 m/s. 

(b) At any instant of time, the velocity of the projectile makes some angle with the horizontal, whose magnitude is obtained as the follows: \[\alpha=\tan^{-1}\left(\frac{v_y}{v_x}\right)\] Substituting the above values into this formula, we get \[\alpha =\tan^{-1}\left(\frac{-28}{8}\right)=-74^\circ\] Therefore, the projectile hits the ground at an angle of 74° below the horizontal. 

Problem (10): From a cliff 100 m high, a ball is kicked at $30^\circ$ above the horizontal with a speed of $20\,{\rm m/s}$. How far from the base of the cliff did the ball hit the ground? (Assume $g=10\,{\rm m/s^2}$).  

Solution: Again, similar to any projectile motion problem, we first select a coordinate system and then draw the path of the projectile as shown in the figure below, 

We choose the origin to be at the kicking point above the cliff, setting $x_0=y_0=0$ in the kinematic equations. 

The coordinate of the hitting point to the ground is $y=-100\,{\rm m}, x = ?$. A negative is inserted because the final point is below the origin.

Now, we find the common quantity in projectile motions—that is, the time between the initial and final points, called total flight time.

To find the total time the ball was in the air, we can use the vertical equation and solve for the unknown $t$ as follows \begin{align*} y&=-\frac 12 gt^2 +(v_0\sin \theta)t \\\\ -100&=-\frac 12 (10) t^2+(20\sin 30^\circ)t\\\\&\Rightarrow \quad \boxed{t^2-2t-20=0} \end{align*} The solutions of a quadratic equation $at^2+bt+c=0$ are found by the formula below \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] By matching the above constant coefficients with the standard quadratic equation, we can find the total time as below \[t=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-20)}}{2(1)}\] After simplifying, two solutions are obtained, $t_1=5.6\,{\rm s}$ and $t_2=-3.6\,{\rm s}$. 

It is obvious that time cannot be negative in physics, so the acceptable answer is $t_1$.

It is the time it takes the ball to travel in the vertical direction. On the other hand, it is also the time it takes the ball to travel the horizontal distance between kicking and hitting points. 

We insert this time into the horizontal equation to find the horizontal distance traveled, known as the projectile's range. \begin{align*} x&=(v_0\cos \theta)t\\ &=(20)(\cos 30^\circ)(5.6)\\&=97\quad {\rm m}\end{align*} 

Problem (11): A cannonball is fired from the top of a $200\,{\rm m}$ cliff with a speed of $60\,{\rm m/s}$ at an angle of $60^\circ$ above the horizontal. Where does the ball land closest to which of the following choices? (Neglect air resistance.) (a) 90m        (b) 402 m        (c) 151 m     (d) 200 m

A projectile from a building

Solution : The horizontal distance from the point where the cannonball fired is wanted, $x=?$. The appropriate kinematic equation for this is \[x=(v_0\cos\theta)t\] The only thing we need is the total time the ball is in the air before landing.

If we take the firing point as the origin, then the cannonball lands $200\,{\rm m}$ below the origin. So, the ball's vertical displacement is $\Delta y=-200\,{\rm m}$. Set this in the vertical displacement equation, $\Delta y=-\frac 12 gt^2+(v_0\sin\theta)t$, and solve for $t$. \begin{gather*} \Delta y=-\frac 12 gt^2 +(v_0\sin\theta)t \\\\ -200=-\frac 12 (10)t^2 +(60\sin 60^\circ)t \\\\  \Rightarrow\quad 5t^2-30\sqrt{3}t-200=0 \\\\ \boxed{t^2-6\sqrt{3}t-40=0}\end{gather*} where we used $\sin 60^\circ=\frac{\sqrt{3}}2$. Using a graphing calculator plot this equation and find its intersections with the horizontal. Consequently, the total flight time is \[\boxed{t=13.4\,{\rm m}}\] Now, substitute this into the horizontal displacement projectile equation below \begin{align*} x&=(v_0\cos\theta)t \\\\ &=(60\,\cos 60^\circ)(13.4) \\\\ &=402\,{\rm m}\end{align*} Thus, the correct answer is (b).

Problem (12): During a kicking drill, a soccer player kicks two successive balls at different angles but with exactly the same speed. The paths that the balls follow are depicted in the figure shown. Which ball remains in the air for a longer time? Neglect air resistance.

Two projectiles at two different angles but the same thrown initial speed.

Solution : both balls land at the same level (or the ground), so the total vertical displacement of both is the same, $\Delta y_1=\Delta y_2=0$. The initial kicking speeds are equal $v_{01}=v_{02}$ and as seen from the figure, ball $1$ has greater kicking angle than ball $2$, i.e., $\theta_1>\theta_2$. Setting $\Delta y=0$ in the following equation and solving for $t$, gives us the total time the ball is in the air \begin{gather*} \Delta y=-\frac 12 gt^2+(v_0 \sin\theta)t \\\\ 0=-\frac 12 gt^2+(v_0 \sin\theta)t \\\\ \Rightarrow \boxed{t=\frac{2v_0 \sin\theta}{g}} \end{gather*} Now, substitute the given data into the equation to compare the flight time of both balls \begin{gather*} \theta_1>\theta_2 \\ \sin\theta_1 > \sin\theta_2 \\\\ \frac{2v_0 \sin\theta_1}{g} > \frac{2v_0 \sin\theta_2}{g} \\\\ \Rightarrow \boxed{t_1>t_2} \end{gather*} Thus, ball $1$ will be in the air for a longer time. 

Author : Dr. Ali Nemati Date Published: 6/15/2021

Updated:  Jun 22, 2023

© 2015 All rights reserved. by Physexams.com

Pediaa.Com

Home » Science » Physics » How to Solve Projectile Motion Problems

How to Solve Projectile Motion Problems

Projectiles are motions involving two dimensions. To solve projectile motion problems, take two directions perpendicular to each other (typically, we use the “horizontal” and the “vertical” directions) and write all vector quantities (displacements, velocities, accelerations) as components along each of these directions. In projectiles,  the vertical motion is independent of the horizontal motion . So, equations of motion can be applied to horizontal and vertical motions separately.

g=9.81\mathrm{\:m\:s^{-1}}

When a projectile thrown at an angle reaches the maximum height, its vertical component of velocity is 0 and when the projectile reaches the same level from which it was thrown, its vertical displacement is 0 . 

How to Solve Projectile Motion Problems

In doing the following calculations, we take upwards direction to be positive in the vertical direction, and horizontally, we take vectors to the right to be positive.

u_v

Strictly speaking, due to air resistance, the path is not parabolic. Rather, the shape becomes more “squashed”, with the particle getting a smaller range.

Proectiles - Effects of Air Resistance

Initially, the object’s vertical speed is decreasing since the Earth is trying to attract it downwards. Eventually, the vertical speed reaches 0. The object has now reached the maximum height. Then, the object starts to move downwards, its downwards velocity increasing as the object is accelerated downwards by gravity.

v_v=0

A person standing at the top of a building 30 m tall throws a rock horizontally from the edge of the building at the speed of 15 m s -1 . Find

a) the time taken by the object to reach the ground,

b) how far away from the building it lands, and

c) the speed of the object when it reaches the ground.

The horizontal velocity of the object does not change, so this isn’t useful by itself to calculate the time. We know the vertical displacement of the object from the top of the building to the ground. If we can find the time taken by the object to reach the ground, we can then find how much the object should move horizontally during that time.

a_v=g=-9.81

A football is kicked off the ground at a speed f 25 m s -1 , with an angle of 20 o to the ground. Assuming there is no air resistance, find how much farther away the ball will land.

u_v=25\mathrm{sin}20^{o}=8.55

About the Author: Nipun

​you may also like these.

Projectile Motion Calculator

Projectile motion step by step.

expand menu

  • Acceleration
  • Centripetal Acceleration
  • Angular Acceleration
  • Impulse (Momentum)
  • Impulse (Velocity)
  • Kinetic Energy
  • Gravitational Potential Energy
  • Projectile Motion
  • Horizontal Projectile Motion

steps for solving projectile motion problems

projectile-motion-calculator

steps for solving projectile motion problems

Related Symbolab blog posts

We want your feedback.

Please add a message.

Message received. Thanks for the feedback.

Generating PDF...

IMAGES

  1. steps in solving projectile motion problems

    steps for solving projectile motion problems

  2. How to Solve Projectile Motion Problems: Applying Newton's Equations of

    steps for solving projectile motion problems

  3. How to Solve Projectile Motion Problems: Applying Newton's Equations of

    steps for solving projectile motion problems

  4. How to Solve Projectile Motion Problems: Applying Newton's Equations of

    steps for solving projectile motion problems

  5. How to Solve a Projectile Motion Problem: 12 Steps (with Pictures)

    steps for solving projectile motion problems

  6. steps in solving projectile motion problems

    steps for solving projectile motion problems

VIDEO

  1. vector addition for projectile motion

  2. || Engineering Mechanics || Projectile Motion || Steps to Solve || [HINDI]

  3. Projectile Motion: Initial Conditions Problems

  4. How To Solve For The Magnitude & Direction Of A Velocity

  5. Projectile Motion Problems: Horizontal Launch

  6. Projectile Motion Problem Solving (5 Examples & Strategies)

COMMENTS

  1. How to Solve a Projectile Motion Problem

    1 Determine what type of problem it is. There are two types of projectile motion problems: (1) an object is thrown off a higher ground than what it will land on. (2) the object starts on the ground, soars through the air, and then lands on the ground some distance away from where it started. 2 Draw a picture.

  2. 5.3 Projectile Motion

    Projectile motion is the motion of an object thrown (projected) into the air when, after the initial force that launches the object, air resistance is negligible and the only other force that object experiences is the force of gravity. The object is called a projectile, and its path is called its trajectory.

  3. Horizontally Launched Projectile Problems

    Problem Type 1: A projectile is launched with an initial horizontal velocity from an elevated position and follows a parabolic path to the ground. Predictable unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight, and the horizontal distance of the projectile.

  4. Projectile Motion

    Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity. To solve projectile motion problems, perform the following steps: 1. Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components.

  5. How to Solve Projectile Motion Problems (Step by Step)

    912 60K views 4 years ago Engineering Dynamics Learn to solve projectile motion problems easily from your textbook step by step. Learn which equations to use, when to use them, and what to do...

  6. 5 Steps to Solve Every Projectile Motion Problem

    Physics tutorial. Use these 5 steps to solve every projectile motion problem. With an example using the 5 steps.00:00 Prerequisites02:02 5 Steps05:54 Example

  7. How to solve a projectile motion problem? (video)

    How to solve a projectile motion problem? What is 2D projectile motion? Projectile at an angle Angled launch projectile vectors Comparing projectile trajectories Projectiles launched at an angle review Horizontally launched projectile Solving kinematic equations for horizontal projectiles Projectile motion graphs

  8. 3.3: Projectile Motion

    I try to go step by step through this difficult problem to layout how to solve it in a super clear way. 2D kinematic problems take time to solve, take notes on the order of how I solved it. Best wishes. ... Introducing the "Toolbox" method of solving projectile motion problems! Here we use kinematic equations and modify with initial ...

  9. How To Solve Any Projectile Motion Problem (The Toolbox Method)

    Introducing the "Toolbox" method of solving projectile motion problems! Here we use kinematic equations and modify with initial conditions to generate a "too...

  10. 4.4: Projectile Motion

    Figure 4.4.2: (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because a x = 0 and v x is a constant. (c) The velocity in the vertical direction begins to decrease as the object rises.

  11. Projectile Motion

    24 Projectile Motion ... Follow the 4 steps of problem solving. You can peak at my solution if you get stuck, but then try to finish up the problem on your own. To end, let's try a more conceptual question (no math needed!). Exercise 24.3: Bullet Fired vs Bullet Dropped.

  12. Steps To Solve Projectile Motion Questions

    Steps To Solve Projectile Motion Questions. September 13, 2015 by Mini Physics. Show/Hide Sub-topics (Kinematics | A Level) Make a clearly labelled sketch of the projectile motion. If you are given the initial velocity, resolve it into its x and y components. Analyze the horizontal and vertical motion separately. Decide on your sign convention.

  13. Ch. 3 Problems & Exercises

    Use graphical methods to solve these problems. You may assume data taken from graphs is accurate to three digits. 1. Find the following for path A in Figure 3.49: (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish.

  14. Projectile Motion Problem Solving

    Projectile Motion Problem Solving It is necessary to understand how to break a vector into its x and y components in order to solve problems for projectiles. Break the Initial Velocity Vector into its Components Apply the Kinematics Equations

  15. Projectile Problems with Solutions and Explanations

    Solution to Problem 1 Problem 2 A projectile is launched from point O at an angle of 22° with an initial velocity of 15 m/s up an incline plane that makes an angle of 10° with the horizontal. The projectile hits the incline plane at point M. a) Find the time it takes for the projectile to hit the incline plane. b)Find the distance OM.

  16. 3.4 Projectile Motion

    Apply the principle of independence of motion to solve projectile motion problems. Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.

  17. 3.4 Projectile Motion

    Figure 3.35 (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because a x = 0 a x = 0 and v x v x is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero.

  18. Grade 12: Physics Worksheet on Projectile Motion

    Solution: The initial step in answering any projectile motion questions is to establish a coordinate system and sketch the path of the projectile, including its initial and final positions and velocities. By doing so, you will be able to solve the relevant projectile equations easily.

  19. Solving Projectile Motion Problems in Physics

    Are you struggling with projectile motion problems in physics? In this video, we'll show you how to solve them step-by-step!We'll start by explaining what pr...

  20. How to Solve Projectile Motion Problems

    To solve projectile motion problems, take two directions perpendicular to each other (typically, we use the "horizontal" and the "vertical" directions) and write all vector quantities (displacements, velocities, accelerations) as components along each of these directions. In projectiles, the vertical motion is independent of the ...

  21. Projectile Motion Problems: Exploring the Physics of Projectiles

    To solve projectile motion problems without initial velocity, follow these steps: Analyze the vertical motion : If the object is launched vertically, you can use the equations of motion for vertical motion to determine the time of flight, maximum height, and other relevant quantities.

  22. College Physics by Openstax Chapter 3 Problem 29

    In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch? Solution:

  23. Projectile Motion Calculator

    You write down problems, solutions and notes to go back... Read More. Free Projectile Motion Calculator - calculate projectile motion step by step.