6.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8e220b8a-46f1-4b9b-88a4-f032c7fbda50-09_323_1018_274_452}
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\caption{Figure 3}
\end{figure}
A child playing cricket on horizontal ground hits the ball towards a fence 10 m away. The ball moves in a vertical plane which is perpendicular to the fence. The ball just passes over the top of the fence, which is 2 m above the ground, as shown in Figure 3.
The ball is modelled as a particle projected with initial speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from point \(O\) on the ground at an angle \(\alpha\) to the ground.
- By writing down expressions for the horizontal and vertical distances, from \(O\) of the ball \(t\) seconds after it was hit, show that
$$2 = 10 \tan \alpha - \frac { 50 g } { u ^ { 2 } \cos ^ { 2 } \alpha }$$
Given that \(\alpha = 45 ^ { \circ }\),
- find the speed of the ball as it passes over the fence.