M2 June 2008 Q7
7.
A ball is thrown from a point \(A\) at a target, which is on horizontal ground. The point \(A\) is 12 m above the point \(O\) on the ground. The ball is thrown from \(A\) with speed 25 m s\(^{-1}\) at an angle of 30\(^\circ\) below the horizontal. The ball is modelled as a particle and the target as a point \(T\). The distance \(OT\) is 15 m. The ball misses the target and hits the ground at the point \(B\), where \(OTB\) is a straight line, as shown in Figure 4. Find
(a) the time taken by the ball to travel from \(A\) to \(B\), (5)
(b) the distance \(TB\). (4)
The point \(X\) is on the path of the ball vertically above \(T\).
(c) Find the speed of the ball at \(X\). (5)
| Scheme | Marks |
|---|---|
| \((\downarrow)\) \(u_y = 25\sin 30^\circ\ \ (= 12.5)\) | B1 |
| \(12 = 12.5t + 4.9t^2\) \(-1\) each error | M1 A2 (1, 0) |
| Leading to \(t = 0.743,\ \ 0.74\) | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \((\rightarrow)\) \(u_x = 25\cos 30^\circ\ \ \left(= \dfrac{25\sqrt{3}}{2} \approx 21.65\right)\) | B1 |
| \(OB = 25\cos 30^\circ \times t\ \ (\approx 16.094\,58)\) ft their (a) | M1 A1ft |
| \(TB \approx 1.1\) (m) awrt 1.09 | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \((\rightarrow)\) \(15 = u_x \times t \Rightarrow t = \dfrac{15}{u_x}\ \ \left(= \dfrac{2\sqrt{3}}{5} \approx 0.693 \text{ or } 0.69\right)\) | M1 A1 |
| either \((\downarrow)\) \(v_y = 12.5 + 9.8t\ \ (\approx 19.2896)\) | M1 |
| \(V^2 = u_x^{\,2} + v_y^{\,2}\ \ (\approx 840.840)\) | |
| \(V \approx 29\) (m s\(^{-1}\)), 29.0 | M1 A1 |
| (5) | |
| (14 marks) |
or
| \((\downarrow)\) \(s_y = 12.5t + 4.9t^2\ \ (\approx 11.0)\) | M1 |
| \(\dfrac{1}{2}m \times 25^2 + mg \times s_y = \dfrac{1}{2}mv^2\) | |
| \(V \approx 29\) (m s\(^{-1}\)), 29.0 | M1A1 |