AP®PhysicsCollege BoardPhysics 1: Algebra-BasedStudy GuidesUnit 1: KinematicsVectors & MotionProjectile Motion

Projectile Motion (College Board AP® Physics 1: Algebra-Based)

Study Guide

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Leander Oates

Written by: Leander Oates

Reviewed by: Caroline Carroll

Projectile motion

  • Projectile motion is a special case of two-dimensional motion

  • Projectile motion will always have:

    • zero acceleration in one dimension

    • constant, nonzero acceleration in the second dimension

  • A projectile is defined as:

An object moving freely under gravity in a two dimensional plane

  • Examples of projectiles include:

    • throwing a ball

    • jumping off a diving board

    • hitting a baseball with a bat

  • Projectiles are usually launched or thrown into the air, at which point the only force acting on the object is the force of weight

    • It is assumed air resistance is ignored

  • Therefore, the constant nonzero acceleration in the second dimension is acceleration due to gravity at the Earth's surface, g space equals space 10 space straight m divided by straight s squared

Calculations involving projectile motion

  • The vertical and horizontal components of motion are independent of one another

  • A projectile will have a parabolic trajectory if it is launched at any angle other than directly vertical

  • The direction of the resultant velocity of a projectile is an angle, θ , to the horizontal

    • The magnitude of the component velocities can be calculated using trigonometry

Diagram of a cannonball's projectile motion, showing angle of projection, resultant velocity, hypotenuse, opposite, and adjacent sides.
An object in a projectile motion trajectory has a resultant velocity at a given angle to the horizontal ground

  • Some key terms to know:

    • Time of flight (total time): how long the projectile is in the air

      • For typical projectile motion, the time to the maximum height is half of the total time

    • Maximum height attained: the height at which the projectile is momentarily at rest

      • This occurs when the vertical velocity component is equal to zero

      • The projectile is at its maximum height when half of its total time has elapsed

    • Range: the horizontal displacement of the projectile

Diagram showing projectile motion with labels for total range (R), initial velocity (u), maximum height (H), and time elapsed. Vertical velocity is 0 at max height.
An object in projectile motion will have a vertical velocity of zero at maximum height when half the time has elapsed

  • For objects in projectile motion:

    • The horizontal velocity is constant

    • Therefore, the horizontal acceleration is zero

    • The vertical velocity changes

    • The vertical acceleration is constant at 10 space straight m divided by straight s squared downward

A diagram showing projectile motion with a cannon firing a ball. It illustrates decreasing and increasing vertical velocity (Vy), constant horizontal velocity (Vx), and constant acceleration (g).
Acceleration and horizontal velocity are always constant whilst vertical velocity changes

Worked Example

A soccer ball at rest on the ground, is kicked with an initial velocity of 12 space straight m divided by straight s at an angle of 35 degree relative to the ground.

Considering the two-dimensional motion of the soccer ball, calculate:

(a) the time of flight.

(b) the maximum height attained.

(c) the range.

Answer:

Analyze the scenario

  • Time of flight is the total time that the ball is in the air

  • The ball reaches its maximum height when its vertical velocity drops to zero

  • Maximum height occurs halfway through its time of flight

  • The range is determined by the horizontal displacement during the time of flight

Part (a) Calculate the time of flight

Step 1: List the known quantities

  • Taking upward as positive

  • Magnitude of resultant initial velocity, v subscript space 0 end subscript space equals space 12 space straight m divided by straight s

  • Direction of initial velocity, theta subscript 0 space equals space 35 degree

  • Acceleration, g space equals space minus 10 space straight m divided by straight s squared

Step 2: Consider the y direction

  • The y component of the initial velocity, v subscript y 0 end subscript, is given by:

v subscript y space 0 end subscript space equals space v subscript 0 space open parentheses sin space theta close parentheses

v subscript y 0 end subscript space equals space 12 space times space sin open parentheses 35 close parentheses

  • At the apex of the projectile motion, the vertical velocity is momentarily zero as it turns from travelling upwards to downwards

    • Use this moment as final velocity in the y direction, v subscript y space equals space 0

Step 3: Apply the correct kinematic equation

  • To find the time taken to reach the apex, use the equation:

v subscript y space equals space v subscript y 0 end subscript space plus space a subscript y t

0 space equals space 12 times sin open parentheses 35 close parentheses space minus g t

g t space equals space 12 times sin open parentheses 35 close parentheses

t space equals space 12 over g sin open parentheses 35 close parentheses

  • The time taken to reach the apex of motion is half the total time

    • This is because the motion is symmetrical

  • Total time of flight is therefore:

t subscript t o t a l end subscript space equals space 2 times 12 over 10 sin open parentheses 35 close parentheses

t subscript t o t a l end subscript space equals space 1.38 space straight s

Part (b) Calculate the maximum height

Step 1: List the known quantities

  • Initial vertical velocity, v subscript y 0 end subscript space equals space 12 sin open parentheses 35 close parentheses

  • Final vertical velocity at apex, v subscript y space equals space 0 space straight m divided by straight s

  • Acceleration, g space equals space minus 10 space straight m divided by straight s squared

  • Total time of flight is known, but it is best to use quantities given in the question

Step 2: Apply the correct fundamental equation

  • To find the displacement at the apex of the trajectory, h, use the following equation and set the initial displacement, y subscript 0, to zero:

v subscript y squared space equals space v subscript y space 0 end subscript superscript 2 space plus space 2 a subscript y open parentheses y space minus space y subscript 0 close parentheses

h space equals space y space minus space y subscript 0

0 space equals space open parentheses 12 times sin open parentheses 35 close parentheses close parentheses squared space minus space 2 g h

  • Rearrange for h and substitute known quantities:

h space equals space fraction numerator open parentheses 12 sin open parentheses 35 close parentheses close parentheses squared over denominator 2 times 10 end fraction

h space equals space 2.37 space straight m

Part (c) Calculate the range

Step 1: List the known quantities

  • Magnitude of resultant initial velocity, v subscript space 0 end subscript space equals space 12 space straight m divided by straight s

  • Direction of initial velocity, theta subscript 0 space equals space 35 degree

  • Total time of flight, t subscript t o t a l end subscript space equals space 1.38 space straight s

Step 2: Consider the horizontal direction

  • In thexdirection, no forces act so acceleration is zero

  • The initial x component of velocity is given by:

v subscript x 0 end subscript space equals space 12 times cos open parentheses 35 close parentheses

  • The soccer ball continues at this horizontal velocity for the duration of the motion

Step 3: Apply the correct kinematic equation

  • To find the horizontal range, R, use the equation:

x space equals space x subscript 0 space plus space v subscript x 0 end subscript t space plus space 1 half a subscript x t squared

  • Acceleration is zero because horizontal velocity is always constant

  • Total time is the same as that of the vertical motion:

x space minus space x subscript 0 space equals space R space equals space 12 times cos open parentheses 35 close parentheses space t subscript t o t a l end subscript space plus space 0

R space equals space 12 times cos open parentheses 35 close parentheses open parentheses 1.38 close parentheses

R space equals space 13.6 space straight m

Examiner Tips and Tricks

The trick to these calculations is setting the relevant velocities to zero. Practice with these techniques prior to your exam.

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    Author
    Reviewer
    Leander Oates

    Author: Leander Oates

    Expertise: Physics

    Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.

    Caroline Carroll

    Author: Caroline Carroll

    Expertise: Physics Subject Lead

    Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.