OxfordAQA International AS and A-level Mathematics: Projectile motion (9660) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section M2.5 (Projectiles) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, which the specification places in International A-level (A2) Unit M2 (Mechanics). Take g = 9.8 m/s² and ignore air resistance. Give answers to 3 significant figures unless told otherwise. Notes: study guide, revision notes, course hub, printable checklist and free diagnostics.

Questions

1. A particle is projected from a point O on level ground with speed 20 m/s at 60° above the horizontal. At time 1.5 s after projection:

(a) find the horizontal and vertical distances of the particle from O. [2] (b) find the speed of the particle and the angle its direction of motion makes with the horizontal. [3]

2. A ball is projected from level ground with speed 35 m/s at an angle α above the horizontal, where sin α = 3/5.

(a) Find the time of flight. [2] (b) Find the greatest height reached. [2] (c) Find the range. [2]

3. A marble rolls off the edge of a horizontal table of height 1.2 m with speed 2.5 m/s and lands on the floor.

(a) Find the time taken to reach the floor. [2] (b) Find the horizontal distance from the edge of the table to the landing point. [1] (c) State one modelling assumption you have made. [1]

4. A particle is projected from the origin O with speed V at angle α above the horizontal. Derive the equation of its path, y = x tan α − g x² sec² α / (2V²), where x and y are horizontal and vertical displacements from O. [4]

5. A ball is projected from a point O on level ground with speed 21 m/s at an angle α above the horizontal. It passes through the point 30 m horizontally from O and 5 m above O.

(a) Show that 2 tan² α − 6 tan α + 3 = 0. [3] (b) Find the two possible values of α, in degrees to 1 decimal place. [3]

6. A particle is projected from the top of a tower 25 m above level ground. The speed of projection is 12 m/s and the direction is 20° below the horizontal.

(a) Find the time taken to reach the ground. [3] (b) Find the horizontal distance from the tower to the landing point. [1] (c) Find the speed of the particle as it reaches the ground. [2]

7. A ball is hit from level ground at 18 m/s at 52° above the horizontal. A net hangs from a horizontal beam 9.5 m above the ground, so the ball can only be stopped by the net if it is higher than 9.5 m. Find the range of horizontal distances from the launch point over which the ball is higher than 9.5 m. [6]

8. A projectile is launched from level ground with speed 21 m/s. Its range is 40 m. Without quoting a range formula, find the two possible angles of projection, to 1 decimal place. [5]

9. (a) State two assumptions made when a tennis ball in flight is modelled as a projectile. [2] (b) Explain why the model gives a longer range than the ball would actually achieve. [2]

10. A footballer kicks a ball from a point O on level ground with speed 25 m/s at an angle α above the horizontal, where tan α = 3/4.

(a) Show that the equation of the path is y = 0.75x − 0.01225x², where x and y are measured in metres. [3] (b) Find the range of the ball on level ground. [2] (c) A vertical wall of height 8 m stands on the level ground 50 m from O, in the plane of the motion. Show that the ball hits the wall, and find how far above the ground it hits. [2] (d) State whether the ball is rising or falling when it hits the wall, and justify your answer. [2]

11. A particle is projected from a point A on the roof of a building, 15 m above level ground, at an angle of 40° above the horizontal with speed V m/s. It lands on the ground 48 m horizontally from A.

(a) Find V. [5] (b) Find the greatest height of the particle above the ground. [2]

Answers

1. (a) Horizontal: x = 20 cos 60° × 1.5 = 10 × 1.5 = 15 m [1]. Vertical: y = 20 sin 60° × 1.5 − 4.9 × 1.5² = 25.98 − 11.025 = 14.96 m [1] (b) Horizontal velocity = 10 m/s; vertical velocity = 17.32 − 9.8 × 1.5 = 2.62 m/s [1]. Speed = √(10² + 2.62²) = 10.3 m/s [1]. Angle = tan⁻¹(2.62 / 10) = 14.7° above the horizontal [1] Examiner insight: Show the two velocity components before combining them, so any arithmetic slip is easy to locate. A direction needs a reference: above or below the horizontal.

2. (a) cos α = 4/5, so vertical component = 35 × 3/5 = 21 m/s and horizontal component = 28 m/s [1]. Time of flight: 21t − 4.9t² = 0, so t = 42/9.8 = 4.29 s [1] (b) The vertical velocity is zero at the top: 0 = 21² − 2 × 9.8 × H [1], so greatest height H = 22.5 m [1] (c) Range = 28 × 42/9.8 [1] = 120 m [1] Examiner insight: With sin α given as a fraction, use the exact components 21 and 28 instead of a decimal angle. This avoids rounding errors in the range.

3. (a) Vertical: initial vertical velocity is zero, so 1.2 = ½ × 9.8 × t² [1]. t² = 0.2449, t = 0.495 s [1] (b) Horizontal distance = 2.5 × 0.4949 = 1.24 m [1] (c) For example: air resistance is negligible (or the marble is a particle) [1] Examiner insight: A horizontal launch means the vertical starting velocity is zero. Say so at the start of the working.

4. Horizontal: x = (V cos α)t, so t = x / (V cos α) [1]. Vertical: y = (V sin α)t − ½gt² [1]. Substitute: y = x tan α − ½g x² / (V² cos² α) [1]. Using 1/cos² α = sec² α: y = x tan α − g x² sec² α / (2V²) [1] Examiner insight: For a derivation the target is given, so every stage must be visible: the expression for t, the substitution and the identities used.

5. (a) Using the trajectory equation with sec² α = 1 + tan² α and T = tan α: 5 = 30T − 4.9 × 900 (1 + T²) / 441 [1]. 4.9 × 900 / 441 = 10, so 5 = 30T − 10 − 10T² [1]. Rearranged: 10T² − 30T + 15 = 0, so 2T² − 6T + 3 = 0 [1] (b) T = (6 ± √12) / 4 = (3 ± √3)/2 [1]. T = 2.366 or 0.634 [1]. α = 67.1° or 32.4° [1] Examiner insight: Both angles are expected, because the point can be reached on the way up or on the way down. Giving only one angle leaves the answer incomplete.

6. (a) Components: horizontal 12 cos 20° = 11.276; vertical 12 sin 20° = 4.104 downwards [1]. Taking up as positive, −25 = −4.104t − 4.9t² [1]. 4.9t² + 4.104t − 25 = 0 gives t = 1.88 s (positive root) [1] (b) Distance = 11.276 × 1.8785 = 21.2 m [1] (c) Vertical velocity = −4.104 − 9.8 × 1.8785 = −22.51 m/s [1]. Speed = √(11.276² + 22.51²) = 25.2 m/s [1] Examiner insight: A launch below the horizontal needs a negative vertical component. Taking it as positive gives a different quadratic and a wrong time.

7. Trajectory: y = x tan 52° − g x² / (2V² cos² 52°) = 1.2799x − 0.03990x² [1]. Put y = 9.5: 0.03990x² − 1.2799x + 9.5 = 0 [1]. Solve the quadratic: x = 11.66 or x = 20.42 [2]. The coefficient of x² is positive here, so the quadratic is negative between the roots, meaning y is above 9.5 m between them [1]. So the ball is higher than 9.5 m for 11.7 m < x < 20.4 m [1] Examiner insight: The answer is an interval, so state both end points and which side is inside. The greatest height, 10.3 m, being above 9.5 m confirms an interval exists.

8. Time of flight: V sin α t − 4.9t² = 0, so t = 2V sin α / g [1]. Range = V cos α × 2V sin α / g = V² sin 2α / g [1]. 40 = 441 sin 2α / 9.8, so sin 2α = 392/441 = 0.8889 [1]. 2α = 62.7° or 117.3° [1]. α = 31.4° or 58.6° [1] Examiner insight: With no formula to quote, the derivation is part of the answer. The second angle is 90° minus the first, a quick check.

9. (a) Any two of: particle model; no air resistance; constant gravity; no wind [2] (b) Air resistance acts against the motion of a real ball [1]. It reduces the speed of the ball, so the real horizontal distance travelled is less than the model gives [1] Examiner insight: An explanation needs a cause and an effect. Writing only “air resistance” does not say which way the range changes.

10. (a) tan α = 3/4, so sec² α = 1 + 9/16 = 25/16 [1]. y = 0.75x − 9.8 × (25/16) x² / (2 × 625) [1]. Coefficient: 9.8 × 25 / (16 × 1250) = 0.01225, so y = 0.75x − 0.01225x² [1] (b) Set y = 0: x(0.75 − 0.01225x) = 0 [1]. x = 0.75/0.01225 = 61.2 m [1] (c) At x = 50: y = 37.5 − 0.01225 × 2500 = 37.5 − 30.625 = 6.875 m [1]. This is less than 8 m, and 50 m is less than the range, so the ball hits the wall at a height of 6.88 m [1] (d) Horizontal speed = 25 × 0.8 = 20 m/s, so the time to reach the wall is 50/20 = 2.5 s, and vertical velocity = 25 × 0.6 − 9.8 × 2.5 = −9.5 m/s [1]. This is negative, so the ball is falling [1] Examiner insight: “Show that” needs each line of the substitution, including the exact sec² α. In part (c) compare the height with 8 m in a sentence.

11. (a) Horizontal: 48 = V cos 40° × t, so t = 48 / (V cos 40°) [1]. Vertical: −15 = V sin 40° × t − 4.9t² [1]. Substitute: −15 = 48 tan 40° − 4.9 × 48² / (V² cos² 40°) [1]. Rearrange: V² = 4.9 × 2304 / (cos² 40° × (48 tan 40° + 15)) = 11289.6 / (0.58682 × 55.278) = 348.0 [1]. V = 18.7 [1] (b) Vertical component = 18.656 × sin 40° = 11.99 m/s, so the rise above A = 11.99² / 19.6 = 7.34 m [1]. Greatest height above the ground = 15 + 7.34 = 22.3 m [1] Examiner insight: The landing level is below the launch point, so y = −15. In part (b) the datum is the ground, not A.

Where marks are usually lost

• Writing V in the equations of motion instead of V cos α and V sin α.
• Using level-ground formulae when the landing point is above or below the launch point.
• Getting the sign of the landing level wrong (y = −25 for a 25 m drop).
• Stating the speed at the top as zero.
• Leaving out the second angle when the quadratic in tan α has two roots.
• Not stating a conclusion when asked whether the ball hits or clears something.

Next steps

• Recap with the revision notes.
• Revisit explanations in the study guide.
• Use the course hub and the printable checklist to plan other topics.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA. Section M2.5: Projectiles.