A small rocket is fired vertically upwards, starting from rest at ground level, and moves with constant acceleration. The rocket reaches a height of 200 m after 10 s.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2018). The mark or mark type is shown in brackets.

1. (i) [M1] Use of s = 1/2(u+v)t: 200 = 1/2 × (0 + v) × 10
2. (i) [A1] v = 40 m s⁻¹ AG (Answer Given)
3. (i) [M1] Use of s = ut + 1/2at² or v = u + at to find 'a' (e.g., 200 = 0 × 10 + 1/2 × a × 10² or 40 = 0 + a × 10)
4. (i) [A1] a = 4 m s⁻²
5. (ii) [M1] Use of v² = u² + 2as with a = -g (e.g., 0 = 40² - 2 × g × s)
6. (ii) [A1] s = 80 m, so maximum height above ground = 200 + 80 = 280 m
7. (iii) [M1] (EITHER) Use of v = u + at to find time to max height from 10s: 0 = 40 - g t₁
8. (iii) [A1] t₁ = 4 s
9. (iii) [M1] Use of s = ut + 1/2at² or s = 1/2at² to find time from max height to ground: 280 = 1/2 g t₂²
10. (iii) [A1] t₂ = √56 ≈ 7.48 s, total time = 10 + t₁ + t₂ = 21.5 s
11. (iii) (M1) (OR) Use of s = ut + 1/2at² from 10s to ground level, with displacement s = -200: -200 = 40t₃ - 1/2 g t₃²
12. (iii) (A1) Correct quadratic for time under gravity: 5t₃² - 40t₃ - 200 = 0 (o.e., e.g. t₃² - 8t₃ - 40 = 0)
13. (iii) (M1) Solution of relevant 3-term quadratic (e.g., t₃ = 4 + √56)
14. (iii) (A1) t₃ ≈ 11.48 s, total time = 10 + t₃ = 21.5 s