7.
\begin{figure}[h]
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\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{7670376b-a7b6-4f8c-83b5-bad3f5ff5162-5_501_1284_425_386}
\end{figure}
At time \(t = 0\) a small package is projected from a point \(B\) which is 2.4 m above a point \(A\) on horizontal ground. The package is projected with speed \(23.75 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle \(\alpha\) to the horizontal, where \(\tan \alpha = \frac { 4 } { 3 }\). The package strikes the ground at the point \(C\), as shown in Fig. 2. The package is modelled as a particle moving freely under gravity.
- Find the time taken for the package to reach \(C\).
A lorry moves along the line \(A C\), approaching \(A\) with constant speed \(18 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). At time \(t = 0\) the rear of the lorry passes \(A\) and the lorry starts to slow down. It comes to rest \(T\) seconds later. The acceleration, \(a \mathrm {~m} \mathrm {~s} ^ { - 2 }\) of the lorry at time \(t\) seconds is given by
$$a = - \frac { 1 } { 4 } t ^ { 2 } , \quad 0 \leq t \leq T .$$
- Find the speed of the lorry at time \(t\) seconds.
- Hence show that \(T = 6\).
- Show that when the package reaches \(C\) it is just under 10 m behind the rear of the moving lorry.