In this question you should take the acceleration due to gravity to be \(10 \text{ms^{-2}\).}
\includegraphics{figure_12}
A small ball \(P\) is projected from a point \(A\) with speed \(39 \text{ms}^{-1}\) at an angle of elevation \(\theta\), where \(\sin \theta = \frac{5}{13}\) and \(\cos \theta = \frac{12}{13}\). Point \(A\) is \(20 \text{m}\) vertically above a point \(B\) on horizontal ground. The ball first lands at a point \(C\) on the horizontal ground (see diagram).
The ball \(P\) is modelled as a particle moving freely under gravity.
- Find the maximum height of \(P\) above the ground during its motion. [3]
The time taken for \(P\) to travel from \(A\) to \(C\) is \(7\) seconds.
- Determine the value of \(T\). [3]
- State one limitation of the model, other than air resistance or the wind, that could affect the answer to part (b). [1]
At the instant that \(P\) is projected, a second small ball \(Q\) is released from rest at \(B\) and moves towards \(C\) along the horizontal ground.
At time \(t\) seconds, where \(t \geq 0\), the velocity \(v \text{ms}^{-1}\) of \(Q\) is given by
$$v = kt^3 + 6t^2 + \frac{3}{2}t,$$
where \(k\) is a positive constant.
- Given that \(P\) and \(Q\) collide at \(C\), determine the acceleration of \(Q\) immediately before this collision. [6]