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First teaching 2023

First exams 2025

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Projectile Motion (SL IB Physics)

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Ashika

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Ashika

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Projectile Motion

What is a projectile?

  • A projectile is a particle moving freely (non-powered), under gravity, in a two-dimensional plane
  • Examples of projectile motion include throwing a ball, jumping off a diving board and hitting a baseball with a baseball bat
  • In these examples, it is assumed that:
    • Resistance from the air or liquid (known as fluid resistance) the object is travelling through is negligible
    • Acceleration due to free-fall, g is constant as the object is moving close to the surface of the Earth

2-6-1-horizontal---vertical-components-diagram-1

Examples of objects in a projectile motion trajectory

  • An object is sent into a projectile motion trajectory with a resultant velocityu at an angleθ to the horizontal
    • Examples of this include a ball thrown from a height and a cannonball launched from a cannon

4-3-4-resultant-velocity-angle-of-projection

An object in a projectile motion trajectory has a resultant velocity at a given angle to the horizontal ground

  • Some key terms to know, and how to calculate them, are:
    • 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 is when the vertical velocity component = 0
      • When the projectile is released and lands on the ground the projectile is at its maximum height when half of its total time has elapsed
    • Range: the horizontal distance travelled by the projectile

4-3-4-max-height

An object in projectile motion will have a vertical velocity of zero at maximum height when half the time has elapsed

Horizontal and Vertical Components

  • The trajectory of an object undergoing projectile motion consists of a vertical component and a horizontal component
    • These quantities are independent of each other
    • Displacement, velocity and acceleration are all vector quantities that are different in both components
    • They need to be evaluated separately using the SUVAT Equations
  Horizontal Component Vertical Component
Displacement
  • Maximum range at the end of the motion when the total time has elapsed
  • Half the range at the maximum height when half the time has elapsed
Maximum height is at the top of the motion when half the time has elapsed
Velocity Constant Zero at maximum height
Acceleration Zero (because velocity remains constant)

Acceleration of free fall, g = 9.8 ms−2  

  • Positive when an object is falling towards Earth
  • Negative when an object is moving away from Earth

1-1-6-projectile-motion-ib-2025-physics

Acceleration and horizontal velocity are always constant whilst vertical velocity changes

  • The resultant velocity of an object in projectile motion can be split into its horizontal and vertical vector components using trigonometry where:
    • Vertical component = opposite side of the projectile triangle
      • opposite = sinθ × hyp = u sinθ
    • Horizontal component = adjacent side of the projectile triangle
      • adjacent = cosθ × hyp = u cosθ

4-3-4-resolving-resultant-velocity-projectile-motion

The resultant velocity at an angle to the horizontal can be resolved using trigonometry into the horizontal and vertical components

  • It can be helpful to see how different equations calculate different quantities using SUVAT equations
  • Examples of obtaining the equations for total time, maximum height and range are shown below

Projectile Motion, downloadable AS & A Level Physics revision notes

Examples of using SUVAT equations to determine the time of flight, maximum height and range of a projectile

Solving problems with projectiles

  • You may be required to calculate the missing quantities from the following projectile motion scenarios:
    • Vertical projection above the horizontal
    • Vertical projection below the horizontal
    • Horizontal projection
    • Projection at an angle, the most common scenario

Worked example

A stone is dropped from the top of a cliff 50.0 m high at an angle of 25.0° below the horizontal. The stone has an initial speed of 30.0 ms−1 and follows a curved trajectory. The stone hits the ground at a horizontal distance from the base of the cliff with a vertical velocity of 33.8 ms−1 .

we-below-vertical-angle

Calculate the distance D

 

Answer:

Step 1: Understand the information given in the question

  • The final vertical velocity of the stone, v = 33.8 ms−1 
  • The horizontal velocity will remain constant throughout the motion
  • The question wants us to calculate the range of the stone
  • The vertical acceleration is +g ms−2

Step 2: Resolve velocity into the vertical and horizontal components

Draw a triangle on the diagram to show the vertical and horizontal velocity components

we-method-step-trig-below-horizontal

Calculate the initial vertical component of velocity, uv using trigonometry:

    • uv = opposite side
    • uv = sinθ × hypotenuse side
    • uv = sin(25) × 30 = 12.68 ms−1
  • Calculate the initial horizontal component of velocity, uH using trigonometry: 
    • uH = adjacent side
    • uH = cosθ × hypotenuse side
    • uH = cos(25) × 30 = 27.19 ms−1

Step 3: Consider the equations of motion in the vertical and horizontal directions

  Vertical Motion Horizontal Motion
u 12.68 ms−1

27.19 ms−1

v 33.8 ms−1

27.19 ms−1

a +9.81 ms−2 0 ms−2
t    
s 50 m m ?

Step 4: Calculate the time of flight from the vertical motion

u + at

33.8 = 12.68 + 9.81t

33.8 - 12.68 = 9.81t

t space equals space fraction numerator 33.8 space minus space 12.68 over denominator 9.81 end fraction

= 2.15 s

Step 5: Calculate the range of the stone, using the elapsed time and horizontal motion

ut

= 27.19 × 2.15

= 58.46 m = 58 m (2 s.f.)

Worked example

A ball is thrown from a point P with an initial velocity u of 12 m s-1 at 50° to the horizontal.

What is the value of the maximum height at Q? (ignoring air resistance)

WE - Projectile Motion Worked Example 3 question image, downloadable AS & A Level Physics revision notes

Answer:

Step 1: Consider the situation

  • In this question, vertical motion only needs to be considered to find the vertical height

Step 2: List the known quantities

  • = 12sin(50) ms−1
  • v = 0 ms−1
  • = −9.81 ms−2
  • s = ?

Step 3: State the correct kinematic equation

v squared space equals space u squared space plus space 2 a s

Step 4: Rearrange the equation to make height, the subject

fraction numerator v squared space minus space u squared over denominator 2 a end fraction space equals space s

Step 5: Substitute in the known quantities and calculate  maximum height, s

s space equals space fraction numerator 0 squared space minus space open parentheses 12 sin open parentheses 50 close parentheses close parentheses squared over denominator 2 space cross times space minus 9.81 end fraction

s space equals space minus 4.3 space straight m

Worked example

A motorcycle stunt-rider moving horizontally takes off from a point 1.25 m above the ground, landing 10 m away as shown.

What was the speed at take-off? (ignoring air resistance)

WE - Projectile Motion Worked Example 2 question image, downloadable AS & A Level Physics revision notes

WE - Projectile Motion Worked Example 2 answer image, downloadable AS & A Level Physics revision notes

Examiner Tip

Make sure you don’t make these common mistakes:

  • Mixing up positive and negative values for vectors
  • Mixing up velocities and distances between horizontal and vertical motion
  • Confusing the direction of sin θ and cos θ
  • Not converting units (mm, cm, km etc.) to metres

Further, it is worth noting that projectile motion is typically symmetrical when air resistance is ignored allowing for use of the peak to find the time of total flight or total horizontal distance by doubling the amount to get from the start point to the peak.

In these exam questions, unless specified, fluid resistance can be ignored

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Ashika

Author: Ashika

Expertise: Physics Project Lead

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.