Use the expressions you found in part (i) to show that the cartesian equation of the trajectory of the ball in terms of \(U\) is
$$y = x \tan 68.5 ^ { \circ } - \frac { 4.9 x ^ { 2 } } { U ^ { 2 } \left( \cos 68.5 ^ { \circ } \right) ^ { 2 } }$$
Use this equation to show again that \(U = 12.0\) (correct to three significant figures).
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{7bcde451-5c86-4ed6-b6f5-62c1ad77618c-3_391_1480_248_364}
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\caption{Fig. 7}
\end{figure}
Fig. 7 shows the graph of \(y = \frac { 1 } { 100 } \left( 100 + 15 x - x ^ { 2 } \right)\).
For \(0 \leqslant x \leqslant 20\), this graph shows the trajectory of a small stone projected from the point Q where \(y \mathrm {~m}\) is the height of the stone above horizontal ground and \(x \mathrm {~m}\) is the horizontal displacement of the stone from O . The stone hits the ground at the point R .