M2 June 2009 Q6
6.
A child playing cricket on horizontal ground hits the ball towards a fence 10 m away. The ball moves in a vertical plane which is perpendicular to the fence. The ball just passes over the top of the fence, which is 2 m above the ground, as shown in Figure 3.
The ball is modelled as a particle projected with initial speed \(u\) m s\(^{-1}\) from point \(O\) on the ground at an angle \(\alpha\) to the ground.
(a) By writing down expressions for the horizontal and vertical distances, from \(O\) of the ball \(t\) seconds after it was hit, show that \[2 = 10\tan\alpha - \frac{50g}{u^2\cos^2\alpha}.\] (6)
Given that \(\alpha = 45^\circ\),
(b) find the speed of the ball as it passes over the fence. (6)
| Scheme | Marks |
|---|---|
| \(\rightarrow\ \ x = u\cos\alpha t = 10\) | M1A1 |
| \(\uparrow\ \ y = u\sin\alpha t - \dfrac{1}{2}gt^2 = 2\) | M1A1 |
| \(\Rightarrow t = \dfrac{10}{u\cos\alpha}\) | |
| \(2 = u\sin\alpha \times \dfrac{10}{u\cos\alpha} - \dfrac{g}{2} \times \dfrac{100}{u^2\cos^2\alpha}\) | M1 |
| \(= 10\tan\alpha - \dfrac{50g}{u^2\cos^2\alpha}\) (given answer) | A1 |
| (6) |
| Scheme | Marks |
|---|---|
| \(2 = 10 \times 1 - \dfrac{100g \times 2}{2u^2 \times 1}\) | M1A1 |
| \(u^2 = \dfrac{100g}{8}\), \(\ u = \sqrt{\dfrac{100g}{8}} = 11.1\) (m s\(^{-1}\)) | A1 |
| \(\dfrac{1}{2}mu^2 = m \times 9.8 \times 2 + \dfrac{1}{2}mv^2\) | M1A1 |
| \(v = 9.1\) m s\(^{-1}\) | A1 |
| (6) | |
| (12 marks) |