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AS · Practice questions · Projectile motion

Split it in two.

Six original Cambridge-style questions on the independence of components, horizontal and angled launches, time of flight, range, and landing speed.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Horizontal and vertical, apart.

01
Analysis
[2 marks]

Explain, in terms of forces, why the horizontal velocity of a projectile stays constant while its vertical velocity changes, assuming air resistance is negligible.

  • There is no horizontal force, so no horizontal acceleration, and the horizontal velocity is constant. ✓
  • Gravity acts vertically downward, giving a constant vertical acceleration g, so the vertical velocity changes throughout. ✓
02
Analysis
[3 marks]

A stone is thrown horizontally at 8.0 m s⁻¹ from a cliff 20 m high. Taking g = 9.8 m s⁻², find the time to reach the sea and the horizontal distance travelled.

  • Vertical: 20 = ½ × 9.8 × t², so t² = 4.08 and t = 2.0 s. ✓
  • Horizontal: x = 8.0 × 2.0 = 16 m. ✓
03
Analysis
[3 marks]

A ball is launched from the ground at 30 m s⁻¹ at 30° above the horizontal. Taking g = 9.8 m s⁻², find its initial horizontal and vertical velocity components and the time of flight.

  • Horizontal: 30 cos 30° = 26 m s⁻¹. Vertical: 30 sin 30° = 15 m s⁻¹. ✓
  • Time of flight: t = 2u sinθ / g = 2 × 15 / 9.8 = 3.1 s. ✓
04
Analysis
[2 marks]

For the ball in question 3, find the horizontal range on level ground.

  • Range = horizontal velocity × time of flight = 26 × 3.1 = 81 m. ✓
  • (Equivalently R = u² sin 2θ / g = 30² × sin 60° / 9.8 ≈ 80 m.) ✓
05
Analysis
[3 marks]

A projectile is fired horizontally at 12 m s⁻¹ from a height and, after 1.5 s, has not yet landed. Taking g = 9.8 m s⁻², find the magnitude of its velocity at that instant.

  • Horizontal velocity stays at 12 m s⁻¹. ✓
  • Vertical velocity = gt = 9.8 × 1.5 = 14.7 m s⁻¹ downward. ✓
  • Speed = √(12² + 14.7²) = √(144 + 216) = √360 ≈ 19 m s⁻¹. ✓
06
Analysis
[2 marks]

A marksman aims a rifle horizontally and, at the exact moment the bullet leaves the barrel, a target is released from rest from the same height some distance away. Ignoring air resistance, explain why the bullet still hits the falling target.

  • Both the bullet and the target start with zero vertical velocity and fall under the same g, so at any instant they have dropped by the same amount. ✓
  • The bullet's horizontal motion carries it to the target's horizontal position in that time, and since both have fallen equally, they meet. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.