A stone is thrown horizontally at 6.0\ m s^-1 and falls for 1.5\ s. Find the horizontal distance travelled. [2 marks]

A projectile is launched at 20\ m s^-1 at 40^. Find the vertical component of its launch velocity. [1 mark]

OCR OCR 2019-style practice: A ball is kicked horizontally at 7.5\ m s^-1 from the top of a cliff 22\ m high. Calculate the time to reach the sea and the horizontal distance travelled. Take g = 9.81\ m s^-2.

Treat the vertical motion as free fall from rest. Using s = (1)/(2)gt^2: 22 = (1)/(2)(9.81)t^2, so t^2 = (44)/(9.81) = 4.49 and t = 2.12\ s. The horizontal velocity is constant, so the range is x = v_x t = 7.5 × 2.12 = 15.9\ m, about 16\ m. Markers reward using the vertical equation with u = 0 to find the time, then using constant horizontal velocity for the distance (about 16\ m).

EnglandPhysicsSyllabus dot point

How do we analyse two-dimensional projectile motion by separating horizontal and vertical components?

Projectile motion: independence of horizontal and vertical motion, constant horizontal velocity and vertical free-fall acceleration, and the effect of air resistance on a real trajectory.

A focused answer to the OCR H556 forces and motion content on projectile motion, covering the independence of horizontal and vertical motion, constant horizontal velocity with vertical free fall, calculating range, height and time of flight, and the effect of air resistance.

Generated by Claude Opus 4.811 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

OCR wants you to analyse projectile motion by treating the horizontal and vertical components independently, recognise that horizontal velocity is constant while the vertical motion is free fall, calculate time of flight, range and maximum height, and describe how air resistance changes a real trajectory.

The answer

Independence of the two directions

The strategy is always the same: resolve into horizontal and vertical components, solve the vertical free-fall problem to find the time, then feed that time into the constant-velocity horizontal motion.

Horizontal motion

Vertical motion

For a launch over level ground, the time of flight is twice the time to the highest point, and the range is $x = v_x \times t_{\text{flight}}$ .

Maximum range launch

A ball is thrown at $18\ \text{m s}^{-1}$ at $30^{\circ}$ above the horizontal over level ground. Find its range. Take $g = 9.81\ \text{m s}^{-2}$ .

step 1: Resolve the velocity

$v_x = 18\cos 30 = 15.6\ \text{m s}^{-1}$ ; $v_y = 18\sin 30 = 9.0\ \text{m s}^{-1}$ .

step 2: Time of flight

Time to the top: $t = \frac{v_y}{g} = \frac{9.0}{9.81} = 0.917\ \text{s}$ . Total flight time $= 2 \times 0.917 = 1.83\ \text{s}$ .

step 3: Range

$x = v_x \times t_{\text{flight}} = 15.6 \times 1.83 = 29\ \text{m}$ .

Examples in context

Ballistics, from basketball free throws to artillery, is projectile motion. Long-jumpers optimise their launch angle and speed to maximise range; on level ground, ignoring drag, the range is greatest at a launch angle of $45^{\circ}$ , because that angle gives the best compromise between time of flight (favoured by a large vertical component) and horizontal speed (favoured by a large horizontal component). Stunt and crash analysis reconstructs a vehicle's launch speed from the distance it travelled off a ramp.

With air resistance the real trajectory of a thrown ball is not a symmetric parabola. Drag acts opposite to the velocity and grows with speed, so it removes more energy on the way up than it returns on the way down: the peak is reached sooner and lower, the descent is steeper than the ascent, and the range is shorter than the drag-free prediction. The optimum launch angle for maximum range also drops below $45^{\circ}$ . This is why golf, javelin and ballistics modelling all treat drag explicitly, and why projectile questions that ignore air resistance always say so in the wording.

Try this

Q1. State the horizontal acceleration of a projectile when air resistance is ignored. [1 mark]

Cue. Zero, so the horizontal velocity is constant.

Q2. A stone is thrown horizontally at $6.0\ \text{m s}^{-1}$ and falls for $1.5\ \text{s}$ . Find the horizontal distance travelled. [2 marks]

Cue. $x = v_x t = 6.0 \times 1.5 = 9.0\ \text{m}$ .

Q3. A projectile is launched at $20\ \text{m s}^{-1}$ at $40^{\circ}$ . Find the vertical component of its launch velocity. [1 mark]

Cue. $v_y = 20\sin 40 = 12.9\ \text{m s}^{-1}$ .

Exam-style practice questions

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

OCR 20195 marksA ball is kicked horizontally at

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

from the top of a cliff

22\ \text{m}

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

g = 9.81\ \text{m s}^{-2}

Show worked answer →

Treat the vertical motion as free fall from rest. Using $s = \frac{1}{2}gt^2$ : $22 = \frac{1}{2}(9.81)t^2$ , so $t^2 = \frac{44}{9.81} = 4.49$ and $t = 2.12\ \text{s}$ .

The horizontal velocity is constant, so the range is $x = v_x t = 7.5 \times 2.12 = 15.9\ \text{m}$ , about $16\ \text{m}$ .

Markers reward using the vertical equation with $u = 0$ to find the time, then using constant horizontal velocity for the distance (about $16\ \text{m}$ ).

OCR 20215 marksA projectile is launched at

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

35^{\circ}

above the horizontal over level ground. Calculate the maximum height reached and the total time of flight. Take

g = 9.81\ \text{m s}^{-2}

Show worked answer →

Resolve the launch velocity: horizontal $v_x = 24\cos 35 = 19.7\ \text{m s}^{-1}$ , vertical $v_y = 24\sin 35 = 13.8\ \text{m s}^{-1}$ .

Maximum height (vertical velocity zero at the top), $v^2 = u^2 - 2gs$ : $0 = 13.8^2 - 2(9.81)s$ , so $s = \frac{190}{19.62} = 9.7\ \text{m}$ .

Time of flight: time up is $t = \frac{v_y}{g} = \frac{13.8}{9.81} = 1.41\ \text{s}$ , doubled by symmetry to $2.8\ \text{s}$ .

Markers reward resolving the velocity, the maximum height about $9.7\ \text{m}$ , and the total flight time about $2.8\ \text{s}$ .

Related dot points

Sources & how we know this

OCR AS and A Level Physics A (H156, H556) specification — OCR (2015)