Pearson Edexcel International Advanced Level Mathematics: Projectiles, vertical motion and kinematics using calculus and vectors (YMA01) – 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 Pearson Edexcel International Advanced Level Mathematics (YMA01), Unit M2 section 1, topics 1.1 to 1.4, from the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification, Issue 3, April 2019. M2 is listed as an optional unit for IAS Further Mathematics and for IAL Mathematics and Further Mathematics. Each question that needs g says to use 9.8 m s⁻². Give answers to 3 significant figures unless told otherwise.

Read the study guide first and keep the revision notes to hand. Also see the course hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. A ball is thrown vertically upwards at 12 m s⁻¹ from a balcony 8 m above horizontal ground. Take g = 9.8.

(a) Find the greatest height of the ball above the ground. [3] (b) Find the time from projection until the ball hits the ground. [3]

2. A stone is thrown vertically upwards from the ground and returns to the ground after 5 s. Take g = 9.8.

(a) Find the speed of projection. [2] (b) Find the greatest height reached. [2] (c) Find the two times at which the stone is 20 m above the ground. [3]

3. A ball A is dropped from rest from the top of a tower 40 m high. At the same instant a ball B is projected vertically upwards from the foot of the tower at 20 m s⁻¹. Take g = 9.8.

(a) Show that the balls meet 2 s after release. [4] (b) Find the height above the ground at which they meet. [1] (c) Find the speed of each ball at the instant they meet. [2]

4. A ball is projected from level ground at 28 m s⁻¹ at 35° above the horizontal. Take g = 9.8.

(a) Find how long the ball is in the air. [3] (b) Find how far it travels horizontally. [2] (c) Find the greatest height above the ground. [2]

5. A ball is thrown horizontally at 14 m s⁻¹ from the top of a tower 44.1 m above level ground. Take g = 9.8.

(a) Find the time taken to reach the ground and the horizontal distance travelled. [3] (b) Find the speed of the ball on landing, and the angle its direction makes with the horizontal. [4]

6. A particle is projected from a point O on level ground at 45° to the horizontal. It passes through a point that is 24 m horizontally and 6 m vertically above O. Find the speed of projection. Take g = 9.8. [6]

7. A particle moves on a straight line. Its displacement from O at time t seconds is x = 2t³ − 15t² + 36t metres, for t ≥ 0.

(a) Find expressions for the velocity and acceleration at time t. [2] (b) Find the times when the particle is instantaneously at rest. [2] (c) Find the total distance travelled in the first 4 s. [4]

8. A particle travels along a line with acceleration (12 − 6t) m s⁻² at time t seconds. At t = 0 the particle is at O with velocity −9 m s⁻¹.

(a) Find the velocity at time t and the displacement from O at time t. [4] (b) Find the total distance travelled in the first 3 s. [2] (c) Find the displacement from O and the speed when t = 5. [2]

9. A particle has position vector r = (t³ − 6t)i + (4t − t²)j metres at time t seconds.

(a) Find the velocity and acceleration at t = 3. [3] (b) Find the speed when t = 3. [1] (c) Find when the particle’s velocity points along j. [2] (d) Find the exact position vector at that time. [2]

10. A particle has acceleration a = (6t − 5)i + (8 − 12t)j m s⁻² at time t seconds. When t = 0, its velocity is (4i + 5j) m s⁻¹ and its position vector is (2i − j) m.

(a) Find the velocity at time t. [4] (b) Find the position vector at time t. [4] (c) Find the distance of the particle from the origin when t = 2. [2]

11. A ball is projected from the top of a tower, 30.4 m above level ground, with speed 15 m s⁻¹ at an angle α above the horizontal, where sin α = 0.8. Take g = 9.8.

(a) Show that the ball hits the ground 4 s after projection. [3] (b) Find the horizontal distance of the landing point from the tower. [2] (c) Find the speed of the ball on landing and the angle its path makes with the horizontal. [4] (d) Find the greatest height of the ball above the ground. [2]

Answers

1. (a) Up positive: 0 = 12² − 2(9.8)s [1], s = 144/19.6 = 7.347 [1]. Greatest height = 8 + 7.35 = 15.3 m [1] (b) s = −8: −8 = 12t − 4.9t² [1], so 4.9t² − 12t − 8 = 0 and t = (12 + √(144 + 156.8))/9.8 [1], giving t = 2.99 s (negative root rejected) [1] Examiner insight: Take the launch point as the origin, so the ground is s = −8. Using s = +8 gives a different quadratic.

2. (a) Time up is 2.5 s and v = 0 there, so 0 = u − 9.8(2.5) [1], u = 24.5 m s⁻¹ [1] (b) H = 24.5² / 19.6 [1] = 30.6 m (30.625) [1] (c) 20 = 24.5t − 4.9t² [1], so 4.9t² − 24.5t + 20 = 0 and t = (24.5 ± √208.25)/9.8 [1], giving t = 1.03 s and t = 3.97 s [1] Examiner insight: “Two times” signals a quadratic with two valid roots.

3. (a) Height of A above the ground = 40 − 4.9t² [1]. Height of B = 20t − 4.9t² [1]. They meet when 40 − 4.9t² = 20t − 4.9t² [1], so 40 = 20t and t = 2 s [1] (b) 20(2) − 4.9(4) = 20.4 m [1] (c) Speed of A = 9.8(2) = 19.6 m s⁻¹ [1]. Speed of B = |20 − 9.8(2)| = 0.4 m s⁻¹ (moving upwards) [1] Examiner insight: A “show that” needs the equation set up and reduced in full. The 4.9t² terms cancelling is the key line to show.

4. (a) Vertical component = 28 sin 35° = 16.06 [1]. 0 = 16.06t − 4.9t² [1], so T = 3.28 s [1] (b) Horizontal component = 28 cos 35° = 22.94 [1]. Range = 22.94 × 3.2776 = 75.2 m [1] (c) H = 16.06² / 19.6 [1] = 13.2 m [1] Examiner insight: Carry unrounded values from part (a) into part (b). Early rounding can shift the third significant figure.

5. (a) 44.1 = 4.9t² [1], so t = 3 s [1]. Distance = 14 × 3 = 42 m [1] (b) Vertical speed = 9.8 × 3 = 29.4 [1]. Speed = √(14² + 29.4²) = √1060.36 [1] = 32.6 m s⁻¹ [1]. tan θ = 29.4/14, so θ = 64.5° below the horizontal [1] Examiner insight: Give the angle with a reference, such as “below the horizontal”. An angle with no reference can lose the last mark.

6. Let w be the horizontal and vertical component of the initial velocity (equal at 45°). Horizontal: 24 = wt [1]. Vertical: 6 = wt − 4.9t² [1]. Substitute: 6 = 24 − 4.9t² [1], so t² = 18/4.9 and t = 1.917 [1]. w = 24/1.917 = 12.52 [1]. Speed = w√2 = 17.7 m s⁻¹ [1] Examiner insight: Write the two component equations and join them through t. A bare final answer leaves nothing to credit if it is wrong.

7. (a) v = 6t² − 30t + 36 [1], a = 12t − 30 [1] (b) 6(t² − 5t + 6) = 0 [1], so t = 2 and t = 3 [1] (c) x(0) = 0, x(2) = 28, x(3) = 27, x(4) = 32 [2]. Distance = 28 + 1 + 5 [1] = 34 m [1] Examiner insight: Split the journey at the turning points. Using x(4) − x(0) = 32 gives displacement, not distance.

8. (a) v = 12t − 3t² + c, and v(0) = −9 gives c = −9 [1], so v = 12t − 3t² − 9 [1]. x = 6t² − t³ − 9t + d with d = 0 [1], so x = 6t² − t³ − 9t [1] (b) v = −3(t − 1)(t − 3), so the particle turns at t = 1, where x = −4. At t = 3, x = 0. Distance = 4 + 4 = 8 m [2] (c) x(5) = 150 − 125 − 45 = −20 m [1]. v(5) = 60 − 75 − 9 = −24, so the speed is 24 m s⁻¹ [1] Examiner insight: Show the constant of integration being fixed from the given condition. Speed is a magnitude, so remove the minus sign from −24 in the final answer.

9. (a) v = (3t² − 6)i + (4 − 2t)j [1], a = 6t i − 2j [1]. At t = 3: v = 21i − 2j and a = 18i − 2j [1] (b) Speed = √(21² + 2²) = √445 = 21.1 m s⁻¹ [1] (c) The i-component of v is zero: 3t² − 6 = 0 [1], so t = √2 s [1] (d) r = (2√2 − 6√2)i + (4√2 − 2)j [1] = −4√2 i + (4√2 − 2)j [1] Examiner insight: Moving parallel to j means the i-component is zero.

10. (a) v = (3t² − 5t + c₁)i + (8t − 6t² + c₂)j [2]. From v(0): c₁ = 4 and c₂ = 5 [1], so v = (3t² − 5t + 4)i + (5 + 8t − 6t²)j [1] (b) r = (t³ − 5t²/2 + 4t + d₁)i + (5t + 4t² − 2t³ + d₂)j [2]. From r(0): d₁ = 2 and d₂ = −1 [1], so r = (t³ − 5t²/2 + 4t + 2)i + (5t + 4t² − 2t³ − 1)j [1] (c) At t = 2: r = 8i + 9j [1]. Distance = √(8² + 9²) = √145 = 12.0 m [1] Examiner insight: Each integration needs its own vector constant, fixed with its own condition.

11. (a) Vertical component = 15 × 0.8 = 12. Landing: s = −30.4, so −30.4 = 12t − 4.9t² [1]. 4.9t² − 12t − 30.4 = 0 [1]. Substituting t = 4 gives 78.4 − 48 − 30.4 = 0, and the other root is negative, so t = 4 s [1] (b) Horizontal component = 15 × 0.6 = 9 [1]. Distance = 9 × 4 = 36 m [1] (c) Vertical velocity = 12 − 9.8(4) = −27.2 [1]. Speed = √(9² + 27.2²) = √820.84 = 28.7 m s⁻¹ [1]. tan θ = 27.2/9 [1], so θ = 71.7° below the horizontal [1] (d) H above the tower = 12²/19.6 = 7.347 [1]. Greatest height above ground = 30.4 + 7.35 = 37.7 m [1] Examiner insight: A “show that” must demonstrate that the stated value satisfies the equation and that the other root is rejected. In part (d), add the tower height.

Where marks are usually lost

• Putting the sign of g the wrong way, so that up is positive but a = +9.8.
• Giving one root of a quadratic in t when both, or the later one, are needed.
• Rounding times or components early, then losing accuracy in the final answer.
• Giving a landing angle with no reference such as “below the horizontal”.
• Reporting displacement when asked for distance, because the turning points were skipped.
• Forgetting the constant of integration, or fixing it from the wrong condition.
• Leaving a vector answer without i and j, or giving speed with a minus sign.

Next steps

• Review the facts and methods in the revision notes.
• Re-read hard sections in the study guide.
• Go to the course hub or print the checklist.
• Take one of the free 10-minute diagnostics at /diagnostics/.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, published by Pearson Education Limited. Unit M2 section 1, topics 1.1 to 1.4.