M2 June 2006 Q5
5. A vertical cliff is 73.5 m high. Two stones \(A\) and \(B\) are projected simultaneously. Stone \(A\) is projected horizontally from the top of the cliff with speed 28 m s\(^{-1}\). Stone \(B\) is projected from the bottom of the cliff with speed 35 m s\(^{-1}\) at an angle \(\alpha\) above the horizontal. The stones move freely under gravity in the same vertical plane and collide in mid-air. By considering the horizontal motion of each stone,
(a) prove that \(\cos\alpha = \tfrac{4}{5}\). (4)
(b) Find the time which elapses between the instant when the stones are projected and the instant when they collide. (4)
| Scheme | Marks |
|---|---|
| \(x_A = 28t\) \(x_B = 35\cos\alpha\ t\) | B1 B1 |
| Meet \(\Rightarrow\ 28t = 35\cos\alpha\, t\ \ \Rightarrow\ \ \cos\alpha = 28/35 = 4/5\ \ *\) | M1 A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(y_A = 73.5 - \tfrac{1}{2}gt^2\) \(y_B = 21t - \tfrac{1}{2}gt^2\) | B1 B1 |
| Meet \(\Rightarrow\ 73.5 = 21t\ \Rightarrow\ t = \underline{3.5\ \text{s}}\) | M1 A1 |
| (4) | |
| (8 marks) |