A particle \(P\) is projected from a fixed point \(O\) on horizontal ground. The particle is projected with speed \(u\) at an angle \(\alpha\) above the horizontal. At the instant when the horizontal distance of \(P\) from \(O\) is \(x\), the vertical distance of \(P\) above the ground is \(y\). The motion of \(P\) is modelled as that of a particle moving freely under gravity.
- Show that \(y = x \tan \alpha - \frac { g x ^ { 2 } } { 2 u ^ { 2 } } \left( 1 + \tan ^ { 2 } \alpha \right)\)
(6)
A small ball is projected from the fixed point \(O\) on horizontal ground. The ball is projected with speed \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at angle \(\theta ^ { \circ }\) above the horizontal. A vertical pole \(A B\), of height 2 m , stands on the ground with \(O A = 10 \mathrm {~m}\), as shown in Figure 3.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0762451f-b951-4d66-9e01-61ecb7b30d95-24_246_899_840_525}
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\caption{Figure 3}
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The ball is modelled as a particle moving freely under gravity and the pole is modelled as a rod.
The path of the ball lies in the vertical plane containing \(O , A\) and \(B\).
Using the model,