M2 June 2010 Q7
7.
A ball is projected with speed 40 m s\(^{-1}\) from a point \(P\) on a cliff above horizontal ground. The point \(O\) on the ground is vertically below \(P\) and \(OP\) is 36 m. The ball is projected at an angle \(\theta^\circ\) to the horizontal. The point \(Q\) is the highest point of the path of the ball and is 12 m above the level of \(P\). The ball moves freely under gravity and hits the ground at the point \(R\), as shown in Figure 3. Find
(a) the value of \(\theta\), (3)
(b) the distance \(OR\), (6)
(c) the speed of the ball as it hits the ground at \(R\). (3)
| Scheme | Marks |
|---|---|
| Vertical motion: \(\quad v^2 = u^2 + 2as\) | M1 |
| \((40\sin\theta)^2 = 2 \times g \times 12\) | A1 |
| \((\sin\theta)^2 = \dfrac{2 \times g \times 12}{40^2}\) | |
| \(\theta = 22.54 = 22.5^\circ\) (accept 23) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| Vert motion \(P \to R:\quad s = ut + \frac{1}{2}at^2\) | |
| \(-36 = 40\sin\theta t - \dfrac{g}{2}t^2\) | M1 |
| \(\dfrac{g}{2}t^2 - 40\sin\theta t - 36 = 0\) | A1 A1 |
| \(t = \dfrac{40\sin 22.54 \pm \sqrt{(40\sin 22.54)^2 + 4 \times 4.9 \times 36}}{9.8}\) | |
| \(t = 4.694\ldots\) | A1 |
| Horizontal P to R: \(\ s = 40\cos\theta t\) | M1 |
| \(= 173\) m (or 170 m) | A1 |
| (6) |
| Scheme | Marks |
|---|---|
| Using Energy: | |
| \(\dfrac{1}{2}mv^2 - \dfrac{1}{2}m \times 40^2 = m \times g \times 36\) | M1 A1 |
| \(v^2 = 2\left(9.8 \times 36 + \frac{1}{2} \times 40^2\right)\) | |
| \(v = 48.0\ldots\) | |
| \(v = 48\) m s\(^{-1}\) (accept 48.0) | A1 |
| (3) | |
| (12 marks) |