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\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a03ad6c1-b4a3-4007-8d3b-ce289a998a55-4_520_582_264_440}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{figure}
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a03ad6c1-b4a3-4007-8d3b-ce289a998a55-4_497_300_287_1411}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{figure}
A small ball \(B\) is projected with speed \(U \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(\theta ^ { \circ }\) above the horizontal from a point \(O\). At time 2 s after the instant of projection, \(B\) strikes a smooth wall which slopes at \(60 ^ { \circ }\) to the horizontal. The speed of \(B\) is \(18 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and its direction of motion is perpendicular to the wall at the instant of impact (see Fig. 1). \(B\) bounces off the wall with speed \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a direction perpendicular to the wall. At time 0.8 s after \(B\) bounces off the wall, \(B\) strikes the wall again at a lower point \(A\) (see Fig. 2).
- Find \(U\) and \(\theta\).
- By considering the motion of \(B\) after it bounces off the wall, calculate \(V\).