A small ball B is projected from a point O on horizontal ground towards a point A 12m above the ground. 0.9 s after projection B has travelled a horizontal distance of 20m and is vertically below A [Figure 1.1].
- (i)Find the angle and the speed of projection of B.[4]
- (ii)Calculate the distance AB when B is vertically below A.[2]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- [M1] (i) tanθ = 12/20
- [A1] (i) θ ( = 30.96) = 31(.0)°
- [M1] (i) Vcos30.96 = 20/0.9
- [A1] (i) V = 25.9 m s⁻¹
- [M1] (ii) H = 25.9sin31 × 0.9 - g x 0.9²/2 ( = 7.948)
- [A1] (ii) AB ( = 12–7.95) = 4.05 m
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →