A ball is kicked at 20 m s^-1 at 30^ above the horizontal. Find its initial vertical velocity component. [1 mark]

AQA AQA 2018-style practice: A ball is thrown horizontally at 12 m s^-1 from a cliff 20 m high. Calculate the time to reach the ground and the horizontal distance travelled. Take g = 9.8 m s^-2.

Treat the vertical and horizontal motions separately, linked by the common time. Vertically the ball starts with zero vertical velocity, so s = 12gt^2 gives 20 = 12(9.8)t^2, so t = 409.8 = 2.0 s. Horizontally the velocity is constant, so the range is 12 × 2.0 = 24 m. Markers reward treating the two motions independently and using the vertical equation to find the time of flight before the horizontal range.

AQA AQA 2021-style practice: A ball is launched from ground level at 25 m s^-1 at an angle of 40^ above the horizontal. Calculate the time of flight and the horizontal range over level ground. Take g = 9.8 m s^-2.

Resolve the launch velocity: u_x = 25cos 40^ = 19.2 m s^-1, u_y = 25sin 40^ = 16.1 m s^-1. For the time of flight over level ground, the ball returns to the same height, so use s = u_y t - 12gt^2 = 0, giving t = 2u_yg = 2(16.1)9.8 = 3.3 s. The range is u_x t = 19.2 × 3.3 = 63 m. Markers reward resolving the velocity, finding the time from the vertical motion returning to zero, and using the constant horizontal velocity for the range.

EnglandPhysicsSyllabus dot point

Why does a projectile follow a curved path, and how do we predict where it lands?

Independence of horizontal and vertical motion, applying the suvat equations to projectiles launched horizontally and at an angle, and the effect of air resistance on the trajectory.

A focused answer to AQA A-Level Physics 3.4.1.4, covering the independence of horizontal and vertical motion, applying the suvat equations to projectiles launched horizontally and at an angle, and the effect of air resistance on the trajectory.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Independence of the two motions
Horizontal launch
Launch at an angle
Air resistance
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What this dot point is asking

AQA specification point 3.4.1.4 wants you to treat horizontal and vertical motion as independent, apply the suvat equations to projectiles launched horizontally and at an angle, and describe qualitatively how air resistance changes the path and range.

Independence of the two motions

This is why a ball dropped and a ball thrown horizontally from the same height hit the ground at the same time: both have the same vertical motion, regardless of their horizontal speed. The horizontal motion neither speeds up nor slows down the fall. The practical consequence is a fixed strategy for every projectile problem: handle the vertical direction with $g$ to find the time of flight, then feed that time into the horizontal direction (constant velocity) to find the range. The only quantity the two directions share is the time, so it acts as the bridge between them.

Horizontal launch

For a projectile launched horizontally with speed $u$ :

The standard method is to find the time of flight from the vertical drop first, then substitute that time into the horizontal equation to get the range. The vertical and horizontal velocities can be combined at any instant to give the resultant velocity, using $v = \sqrt{v_x^2 + v_y^2}$ .

Launch at an angle

For a launch speed $u$ at angle $\theta$ above the horizontal, resolve the velocity into components before applying suvat to each direction.

Projectile launched at an angle over level ground

A stone is launched at $20 \text{ m s}^{-1}$ at $30^{\circ}$ above the horizontal from level ground. Find the maximum height and the range. Take $g = 9.8 \text{ m s}^{-2}$ .

step 1: Resolve the initial velocity

$u_x = u\cos\theta = 20\cos 30^{\circ} = 17.3 \text{ m s}^{-1}$ ; $u_y = u\sin\theta = 20\sin 30^{\circ} = 10.0 \text{ m s}^{-1}$ .

step 2: Find the maximum height

At the top the vertical velocity is zero, so $v^2 = u_y^2 - 2gs$ gives $0 = (10.0)^2 - 2(9.8)s$ , so $s = \dfrac{100}{19.6} = 5.1 \text{ m}$ .

step 3: Find the time of flight

The vertical motion returns to zero, so $t = \dfrac{2u_y}{g} = \dfrac{2(10.0)}{9.8} = 2.04 \text{ s}$ .

step 4: Find the range

Using the constant horizontal velocity, range $= u_x t = 17.3 \times 2.04 = 35 \text{ m}$ .

Air resistance

In reality air resistance opposes motion, so it removes kinetic energy. The range and maximum height are reduced, and the trajectory is no longer a symmetric parabola: the descent is steeper than the ascent, because the horizontal velocity is continuously eroded. The optimum launch angle for maximum range falls below the ideal $45^{\circ}$ once air resistance is significant.

Try this

Q1. State why the horizontal and vertical motions of a projectile can be treated separately. [1 mark]

Cue. Gravity acts vertically only, so it does not affect the horizontal velocity.

Q2. A ball is kicked at $20 \text{ m s}^{-1}$ at $30^{\circ}$ above the horizontal. Find its initial vertical velocity component. [1 mark]

Cue. $u_y = 20\sin 30^{\circ} = 10 \text{ m s}^{-1}$ .

Q3. State the effect of air resistance on the range of a projectile. [1 mark]

Cue. It reduces the range.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20184 marksA ball is thrown horizontally at

12 \text{ m s}^{-1}

from a cliff

20 \text{ m}

high. Calculate the time to reach the ground and the horizontal distance travelled. Take

g = 9.8 \text{ m s}^{-2}

Show worked answer →

Treat the vertical and horizontal motions separately, linked by the common time.

Vertically the ball starts with zero vertical velocity, so $s = \tfrac{1}{2}gt^2$ gives $20 = \tfrac{1}{2}(9.8)t^2$ , so $t = \sqrt{\dfrac{40}{9.8}} = 2.0 \text{ s}$ .

Horizontally the velocity is constant, so the range is $12 \times 2.0 = 24 \text{ m}$ .

Markers reward treating the two motions independently and using the vertical equation to find the time of flight before the horizontal range.

AQA 20215 marksA ball is launched from ground level at

25 \text{ m s}^{-1}

at an angle of

40^{\circ}

above the horizontal. Calculate the time of flight and the horizontal range over level ground. Take

g = 9.8 \text{ m s}^{-2}

Show worked answer →

Resolve the launch velocity: $u_x = 25\cos 40^{\circ} = 19.2 \text{ m s}^{-1}$ , $u_y = 25\sin 40^{\circ} = 16.1 \text{ m s}^{-1}$ .

For the time of flight over level ground, the ball returns to the same height, so use $s = u_y t - \tfrac{1}{2}gt^2 = 0$ , giving $t = \dfrac{2u_y}{g} = \dfrac{2(16.1)}{9.8} = 3.3 \text{ s}$ .

The range is $u_x t = 19.2 \times 3.3 = 63 \text{ m}$ .

Markers reward resolving the velocity, finding the time from the vertical motion returning to zero, and using the constant horizontal velocity for the range.

Related dot points

Sources & how we know this

AQA A-level Physics (7408) specification — AQA (2017)