The problems are on the projectile motion worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Projectile motion combines a constant horizontal velocity with a vertical motion of constant acceleration g = 9.81 m/s² downwards. These worked answers correct the 20 problems of the Physics 11-12 worksheet on projectile motion, from true or false to launch from height and range: each solution shows the formula, the substitution and the final answer with its unit.
Each answer separates the horizontal and the vertical motion and shows the formula, the substitution and the result in bold. Intermediate values are kept unrounded, so your last digit may differ slightly. Compare your method as well as your numbers.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.4
Problem 1 True or false
1. True: no horizontal force acts.2. True: the vertical component changes sign there.
3. False: the acceleration is 9.81 m/s² downwards everywhere.
4. False: mass does not affect the motion without air resistance.
Problem 2 Launch components
1. 35 cos 25° = 35 × 0.906 = 31.7 m/s.2. 35 sin 25° = 35 × 0.423 = 14.8 m/s.
Problem 3 Stone from a bridge
1. t = √(2 × 20 ÷ 9.81) = √4.077 = 2.02 s.2. 2.02 s: the vertical motion starts from zero vertical velocity for both stones and does not depend on the horizontal velocity.
Problem 4 Garden hose
1. t = √(2 × 0.80 ÷ 9.81) = √0.1631 = 0.40 s.2. x = 4.0 × 0.404 = 1.6 m.
Problem 5 Strobe picture
1. At every flash the two balls are at the same height, so their vertical motions are identical.2. The horizontal spacing is equal, so the horizontal velocity is constant.
Problem 6 Vertical component and flight time
1. T = 2uy ÷ g = 29.4 ÷ 9.81 = 3.0 s.2. H = uy² ÷ 2g = 216.1 ÷ 19.62 = 11.0 m.
Problem 7 A six in cricket
1. ux = 28 cos 35° = 22.9 m/s; uy = 28 sin 35° = 16.1 m/s.2. T = 2 × 16.06 ÷ 9.81 = 3.27 s.
3. R = 22.94 × 3.274 = 75.1 m, which is more than 70 m, so the ball lands beyond the rope.
Problem 8 Sprinkler jet
1. t = √(3.6 ÷ 9.81) = 0.61 s.2. x = 6.0 × 0.606 = 3.6 m.
3. vy = 9.81 × 0.606 = 5.94 m/s; v = √(6.0² + 5.94²) = 8.4 m/s; angle = tan⁻¹(5.94 ÷ 6.0) = 44.7°.
Problem 9 Kick from a wall
1. t = √(4.0 ÷ 9.81) = 0.64 s.2. u = 6.0 ÷ 0.6386 = 9.4 m/s.
Problem 10 Complementary angles
1. R = u² sin 2θ ÷ g = 625 × sin 60° ÷ 9.81 = 55.2 m.2. 60°, because 30° + 60° = 90°.
3. The 60° throw goes higher and stays in the air longer.
Problem 11 Maximum range
1. At 45°, R = 38² ÷ 9.81 = 1444 ÷ 9.81 = 147 m.2. R = 147.2 × sin 60° = 127 m.
3. (147.2 − 127.5) ÷ 147.2 = 13 %.
Problem 12 Effect of drag
1. The range is shorter and the maximum height is lower (the path is not symmetrical).2. Drag removes horizontal velocity all the way, while gravity keeps adding downward velocity, so the horizontal part is smaller on the way down.
Problem 13 Archery
1. t = 30 ÷ 60 = 0.50 s.2. Drop = ½ × 9.81 × 0.25 = 1.2 m.
3. Aim about 1.2 m above the target centre.
Problem 14 Velocity after one second
1. ux = 22 cos 50° = 14.1 m/s. vy = 22 sin 50° − 9.81 × 1.0 = 16.85 − 9.81 = 7.04 m/s.2. v = √(14.14² + 7.04²) = 15.8 m/s.
3. tan⁻¹(7.04 ÷ 14.14) = 26.5°.
Problem 15 Over the crossbar
1. ux = 18 cos 40° = 13.79 m/s, so t = 20 ÷ 13.79 = 1.45 s.2. y = 11.57 × 1.4505 − 4.905 × 1.4505² = 16.78 − 10.32 = 6.5 m.
3. 6.5 m is greater than 2.44 m, so the ball passes over the bar.
Problem 16 Ball against a wall
1. ux = 7.5 m/s, so t = 8.0 ÷ 7.5 = 1.07 s.2. y = 12.99 × 1.0667 − 4.905 × 1.0667² = 13.86 − 5.58 = 8.3 m.
3. vy = 12.99 − 9.81 × 1.0667 = +2.5 m/s, so the ball is still rising.
Problem 17 Ball from a cliff
1. uy = 10.0 m/s, ux = 17.32 m/s. −30 = 10t − 4.905t²: t = (10 + √688.6) ÷ 9.81 = 3.69 s.2. x = 17.32 × 3.694 = 64.0 m.
3. v² = 20² + 2 × 9.81 × 30 = 988.6, so v = 31.4 m/s.
4. vy = 10 − 9.81 × 3.694 = −26.2 m/s, so θ = tan⁻¹(26.2 ÷ 17.32) = 56.6°.
Problem 18 Finding the launch angle
1. sin 2θ = gR ÷ u² = 9.81 × 30 ÷ 400 = 0.7358, so 2θ = 47.4° or 132.6°, giving θ = 23.7° or 66.3°.2. T = 2u sin θ ÷ g: 1.64 s for 23.7° and 3.73 s for 66.3°.
Problem 19 Motorbike jump
1. R = 22² × sin 50° ÷ 9.81 = 484 × 0.766 ÷ 9.81 = 37.8 m.2. 37.8 m is more than 35 m, so the rider clears the gap.
3. u² = gR ÷ sin 2θ = 343.4 ÷ 0.766 = 448.2, so u = 21.2 m/s.
Problem 20 Launch from height and range
1. H = u² sin²θ ÷ 2g and R = 2u² sinθ cosθ ÷ g, so H ÷ R = tanθ ÷ 4 and tan θ = 4H ÷ R.2. tan θ = 72 ÷ 48 = 1.5, so θ = 56.3°.
3. u² = gR ÷ sin 2θ = 470.9 ÷ 0.9231 = 510.1, so u = 22.6 m/s.
