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How do you analyse the motion of a projectile by treating its horizontal and vertical motion separately?

Motion of a projectile under gravity, resolving the initial velocity into horizontal and vertical components, the independence of horizontal and vertical motion, and finding range, maximum height and time of flight.

A focused answer to the OCR A-Level Mathematics A projectiles content, covering motion under gravity in two dimensions, resolving the initial velocity into horizontal and vertical components, the independence of the two motions, and finding the time of flight, range, maximum height and the equation of the trajectory.

Generated by Claude Opus 4.812 min answer

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

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

OCR wants you to model a projectile moving freely under gravity, resolve the initial velocity into horizontal and vertical components, treat the two motions independently (constant horizontal velocity, constant vertical acceleration gg), and find the time of flight, horizontal range, greatest height, the velocity at a given time, and the equation of the path.

The answer

The key idea: independent motions

A projectile's horizontal and vertical motions are completely independent and are linked only by the common time. Horizontally there is no force (ignoring air resistance), so the horizontal velocity is constant. Vertically the only force is gravity, so the vertical motion is constant-acceleration motion with a=ga = g downward.

Time of flight, range and height

The vertical motion sets the timing; the horizontal motion then gives the distance.

The velocity at a moment

The horizontal component stays uxu_x; the vertical component is vy=uy−gtv_y = u_y - gt. Combine them for the speed vx2+vy2\sqrt{v_x^2 + v_y^2} and the direction tan⁡−1 ⁣(vyvx)\tan^{-1}\!\left(\dfrac{v_y}{v_x}\right).

Examples in context

A horizontal launch from a height

When a projectile is launched horizontally, the initial vertical velocity is zero, so the fall time comes from the height alone, and the horizontal distance follows.

The equation of the trajectory

Eliminating tt between the horizontal and vertical equations gives yy as a function of xx, a downward parabola. This is useful when a question asks whether a projectile clears an obstacle at a known horizontal distance.

Try this

Q1. A projectile is launched at 3030 m s−1^{-1} and 60∘60^\circ. Find the vertical component of its initial velocity. [1 mark]

  • Cue. uy=30sin⁡60∘=30×32≈26.0u_y = 30\sin 60^\circ = 30 \times \tfrac{\sqrt{3}}{2} \approx 26.0 m s−1^{-1}.

Q2. A stone is thrown horizontally at 88 m s−1^{-1} and falls for 22 s. Find the horizontal distance travelled. [2 marks]

  • Cue. Horizontal velocity is constant, so x=8×2=16x = 8 \times 2 = 16 m.

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 20197 marksA ball is projected from ground level with speed 2525 m s−1^{-1} at 40∘40^\circ above the horizontal. Taking g=9.8g = 9.8 m s−2^{-2} and ignoring air resistance, find the time of flight, the horizontal range, and the greatest height.
Show worked answer →

Resolve the initial velocity (M1): horizontal ux=25cos⁡40∘≈19.15u_x = 25\cos 40^\circ \approx 19.15 m s−1^{-1}, vertical uy=25sin⁡40∘≈16.07u_y = 25\sin 40^\circ \approx 16.07 m s−1^{-1}.

Time of flight: vertical displacement returns to zero, 0=uyt−12gt20 = u_y t - \tfrac{1}{2}gt^2, so t=2uyg=2(16.07)9.8≈3.28t = \dfrac{2u_y}{g} = \dfrac{2(16.07)}{9.8} \approx 3.28 s (M1, A1).

Range: horizontal distance =uxt=19.15×3.28≈62.8= u_x t = 19.15 \times 3.28 \approx 62.8 m (M1, A1).

Greatest height: at the top vy=0v_y = 0, so uy2=2gHu_y^2 = 2gH gives H=16.0722(9.8)≈13.2H = \dfrac{16.07^2}{2(9.8)} \approx 13.2 m (M1, A1).

Markers reward resolving the velocity, the time of flight from vertical motion, the range from horizontal motion, and the greatest height.

OCR 20216 marksA stone is thrown horizontally at 1212 m s−1^{-1} from the top of a cliff 3030 m above the sea. Taking g=9.8g = 9.8 m s−2^{-2}, find the time to reach the sea and the horizontal distance travelled, and the speed of the stone as it hits the sea.
Show worked answer →

Horizontal velocity is constant at 1212 m s−1^{-1}; initial vertical velocity is 00 (B1).

Vertical motion to fall 3030 m: 30=12(9.8)t230 = \tfrac{1}{2}(9.8)t^2, so t2=609.8=6.122t^2 = \dfrac{60}{9.8} = 6.122 and t≈2.47t \approx 2.47 s (M1, A1).

Horizontal distance =12×2.47≈29.7= 12 \times 2.47 \approx 29.7 m (A1).

Vertical speed on landing: vy=gt=9.8×2.47≈24.2v_y = gt = 9.8 \times 2.47 \approx 24.2 m s−1^{-1}. Speed =122+24.22=144+585.6≈27.0= \sqrt{12^2 + 24.2^2} = \sqrt{144 + 585.6} \approx 27.0 m s−1^{-1} (M1, A1).

Markers reward the constant horizontal velocity, the fall time, the horizontal distance, and combining components for the landing speed.

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