A particle is projected vertically upwards with speed 40 m s⁻¹ alongside a building of height h m.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.

1. [4] (a) 40 - gt = 0 [t=4]
2. [4] (a) Using v = u + at with u = 40, v = 0 and a = –g to find the time taken to reach the highest point.
3. [4] (a) Time to top of building = 4 – ½(4) = 2
4. [4] (a) May see t = 4 + 2 = 6 for A1
5. [4] (a) h = 40 × 2 - ½ × 10 × 2²
6. [4] (a) h = 40 × 6 - ½ × 10 × 6²
7. [4] (a) Using s = ut + ½ at² with u = 40, a = -g and t = 2 or t = 6 to set up an equation which enables the value of h, the height of the building, to be found.
8. [4] (a) h = 60
9. [4] (a) Alternative method for question 5(a)
10. [4] (a) 0 = 40² + 2 × (-10) × H
11. [4] (a) For using v² = u² + 2as with u = 40, v = 0 and a = –g in order to find H, the greatest height achieved
12. [4] (a) H = 80
13. [4] (a) s = ½ × 10 × 2²
14. [4] (a) Use either s = vt – ½ at² with v = 0, a = -g, t = 2 or use s = ut + ½ at² with u = 0, a = g, t = 2 to find the distance travelled either in the final 2 seconds going up or the first 2 seconds going down
15. [4] (a) s = 20 and so h = 80 - 20 = 60
16. [4] (b) Height of first particle above ground = 40t – ½ × 10t²
17. [4] (b) Height of second particle above top of building = 20(t − 1) – ½ × 10 × (t − 1)²
18. [4] (b) 60 + 20(t-1) - ½ × 10 × (t − 1)² = 40t – ½ × 10t²
19. [4] (b) Set up an equation involving expressions for displacement to enable the time at which the particles reach the same height to be found.
20. [4] (b) t = 3.5 seconds
21. [4] (b) Alternative method for question 5(b)
22. [4] (b) h₁ = 40 × 1 – 5 × 1² [= 35] and v₁ = 40 – 10 × 1 [= 30]
23. [4] (b) Distance travelled and speed of first particle after 1 second
24. [4] (b) H₁ = 30T – 5 × T², H₂ = 20T – 5 × T²
25. [4] (b) Distance travelled by both particles, T seconds after the second particle is projected.
26. [4] (b) 30T – 5 × T² = 20T – 5 × T² + (60 – 35)
27. [4] (b) Set up an equation in T involving expressions for displacement to enable the time at which the particles are at the same height to be found.
28. [4] (b) T = 2.5 and so time to meet = 2.5 + 1 = 3.5 seconds