ASPhysicsAQARevision NotesMechanics & MaterialsEquations of MotionProjectile Motion

Projectile Motion (AQA AS Physics): Revision Note

Exam code: 7407

Written by: Katie M

Reviewed by: Tim

Updated on 8 October 2026

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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

An archer’s arrow follows a curved path to a target. A football is kicked along a parabolic path over the ground. An angry bird is projected from a building at a pig on the ground. — **Examples of objects in a projectile motion trajectory**

An object is sent into a projectile motion trajectory with a resultant velocity, u at an angle, $θ$ to the horizontal
- Examples of this include a ball thrown from a height and a cannonball launched from a cannon

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

Projectile trajectory showing maximum height H at half the total time, where the vertical velocity is zero. The projectile lands after total time t, having travelled horizontal range R. — **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 m s⁻² Positive when an object is falling towards Earth Negative when an object is moving away from Earth

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 m s⁻²

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

A cannonball’s parabolic trajectory shows vertical velocity V_v decreasing to zero at maximum height, then increasing and changes direction. Horizontal velocity V_h and downward acceleration g remain constant through trajectory. — **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

A cannon launches a projectile at resultant velocity u and angle theta to the ground. u is the hypotenuse, with vertical opposite and horizontal adjacent components. The trajectory reaches a target 250 metres away; vᵧ = u sin θ and uₓ = u cos θ. — **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

Worked derivation for projectile motion where initial velocity components are u sin theta vertically and u cos theta horizontally. SUVAT equations give total time of flight equal to 2u sin theta divided by g, maximum height reached equals (u sin theta) squared divided by 2g, and horizontal range equals u squared sin (2theta) divided by g. — **Examples of using SUVAT equations to determine the time of flight, maximum height and range of a projectile**

The effects of air resistance on projectile motion

Any object moving through the air experiences air resistance which creates a force of drag
This is caused by friction between the air molecules and the object
The drag force acts in the opposite direction to the direction of motion of the object
- Its horizontal component reduces its horizontal velocity and its range
- Its vertical component reduces its maximum height and causes a steeper gradient as it falls back to earth

Factors that affect the motion of an object through a fluid

A fluid is a term used to describe a liquid and a gas
The mass and surface area of an object are the two main factors that affect the motion of an object:
- An object with a larger surface area will experience a larger air resistance
- An object with a greater mass will have a greater weight

Solving problems with projectiles

Questions may ask for the missing quantities in the following projectile motion scenarios:
- Vertical projection
- Horizontal projection
- Projection at an angle, the most common scenario

Worked Example

To calculate vertical projection (free fall)

A science museum designed an experiment to show the fall of a feather in a vertical glass vacuum tube.

The time of fall from rest is 0.5 s.

A feather falling down a vertical vacuum tube of length L.

What is the length of the tube, L?

[2]

Answer:

Step 1: List the known quantities

Length of the tube, $s = L$
Initial velocity, $u = 50 m s^{- 1}$
Acceleration, $a = 9.81 m s^{- 2}$
Time, $t = 0.5 s$

Step 2: Choose the equation which contains the quantities listed in Step 1

The equation that links $s$ , $u$ , $a$ and $t$ is

$s = u t + \frac{1}{2} a t^{2}$

Step 3: Calculate the length

$L = 0 \times 0.5 + \frac{1}{2} \times 9.8 \times 0 . 5^{2}$ [1 mark]

$L = 1.2 m$ [1 mark]

Worked Example

To calculate horizontal projection

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?

A motorcycle taking off horizontally from a point 1.25 m above the ground and landing 10 m away. A dashed curved line shows the projectile trajectory.

[3]

Answer:

Step 1: Consider the vertical motion first and list the known quantities:

Height, $s = 1.25 m$
Initial velocity, $u = 0$
Acceleration, $a = 9.81 m s^{- 2}$
Time of flight, $t$

Step 2: Calculate the time of flight using the quantities listed in Step 1

The equation that links $s$ , $u$ , $a$ and $t$ is

$s = u t + \frac{1}{2} a t^{2}$

$s = \frac{1}{2} a t^{2}$

$t = \sqrt{\frac{2 s}{a}} = \sqrt{\frac{2 \times 1.25}{9.81}} = 0.505 s$ [1 mark]

Step 3: Consider the horizontal motion and list the known quantities:

Distance travelled, $s = 10 m$
Initial velocity, $u$
Acceleration, $a = 0$
Time of flight, $t = 0.505 s$

Step 4: Calculate the speed at take-off

Since the acceleration is zero, use the following equation

$velocity = \frac{displacement}{time}$

$u = \frac{10}{0.505}$ [1 mark]

$u = 19.8 m s^{- 1}$ [1 mark]

Worked Example

To calculate projection at an angle

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?

A ball thrown from point P at 12 m s-1 and 50° to the horizontal, following a curved path to point Q at its maximum height. The ground line is labelled horizontal.

[3]

Answer:

Step 1: Consider the vertical motion and list the known quantities:

Maximum height, $s$
Final velocity, $v = 0$
Acceleration, $a = - 9.81 m s^{- 2}$

Step 2: Calculate the vertical component of velocity

$vertical velocity = \sin θ \times resultant velocity$

$u = 12 \sin (50)$ [1 mark]

Step 2: Calculate the maximum height using the quantities listed in Step 1

The equation that links $s$ , $u$ , $v$ and $a$ is

$v^{2} = u^{2} + 2 a s$

$2 a s = v^{2} - u^{2}$

$s = \frac{v^{2} - u^{2}}{2 a} = \frac{0 - {(12 \sin (50))}^{2}}{2 \times (- 9.81)}$ [1 mark]

$s = 4.3 m$ [1 mark]

Examiner Tips and Tricks

Make sure you remember the following:

Deceleration is negative as the object rises
sin θ is used for the vertical component of velocity and cos θ the horizontal component
Units should be converted into standard units such as m, s, m s⁻¹ and m s⁻²
At the top of the motion, at maximum height, the vertical velocity is zero

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Projectile Motion (AQA AS Physics): Revision Note

Projectile motion

What is a projectile?

Horizontal and vertical components

The effects of air resistance on projectile motion

Factors that affect the motion of an object through a fluid

Solving problems with projectiles

Worked Example

Worked Example

Worked Example

Examiner Tips and Tricks

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Katie M

Reviewer: Tim