6 A section of a golf practice ground is inclined at \(15 ^ { \circ }\) to the horizontal. A golfer is hitting a ball up and down a line of greatest slope of this section of the practice ground. The golfer hits the ball up the slope, so that the ball initially makes an angle of \(30 ^ { \circ }\) with the slope. The ball first bounces on the slope 50 m from its point of projection.

1. Determine the initial speed of the ball. The golfer now hits the ball down the slope. The ball initially moves with speed \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the ball initially travels at an angle \(\theta\) above the horizontal, as shown in Fig. 6. The ball first bounces at a point a distance \(L\) m down the slope. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{37798594-8cb0-48aa-8401-090f09e25dff-6_545_791_794_242} \captionsetup{labelformat=empty} \caption{Fig. 6}
   \end{figure}
2. Show that \(\mathrm { L } = \frac { 800 } { \mathrm {~g} } \left( \frac { \sin \theta \cos \theta } { \cos 15 ^ { \circ } } + \frac { \sin 15 ^ { \circ } \cos ^ { 2 } \theta } { \cos ^ { 2 } 15 ^ { \circ } } \right)\). You are given that \(\frac { \mathrm { dL } } { \mathrm { d } \theta } = \frac { 800 } { \mathrm {~g} } \left( \frac { \cos 2 \theta } { \cos 15 ^ { \circ } } - \frac { \sin 15 ^ { \circ } \sin 2 \theta } { \cos ^ { 2 } 15 ^ { \circ } } \right)\).
3. Determine the value of \(\theta\) for which \(\frac { \mathrm { d } L } { \mathrm {~d} \theta } = 0\).
4. Hence determine the maximum distance the golfer can hit the ball down the slope.