7 In this question take \(\boldsymbol { g } = \mathbf { 1 0 }\).
A small stone is projected from a point O with a speed of \(26 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle \(\theta\) above the horizontal. The initial velocity and part of the path of the stone are shown in Fig. 7.
You are given that \(\sin \theta = \frac { 12 } { 13 }\).
After \(t\) seconds the horizontal displacement of the stone from O is \(x\) metres and the vertical displacement is \(y\) metres.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{ff44367e-c992-4e79-b255-5a04e0b8e21e-07_419_479_904_248}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{figure}
- Using the standard model for projectile motion,
- show that \(y = 24 t - 5 t ^ { 2 }\),
- find an expression for \(x\) in terms of \(t\).
The stone passes through a point A . Point A is 16 m above the level of O . - Find the two possible horizontal distances of A from O .
A toy balloon is projected from O with the same initial velocity as the small stone.
- Suggest two ways in which the standard model could be adapted.