91Ó°ÊÓ

The range R of the ball as a function is $R\left(\mathrm{θ}\right)=672\sin \left(2\mathrm{θ}\right)$, where Ris measured in feet and the initial velocity is $100$miles per hour.

We have to find the angle of the ball if the golfer wants the ball to travel450feet(150yards).

By using the function of the ball we get,

$$\begin{gathered} R\left(\mathrm{θ}\right)=672\sin \left(2\mathrm{θ}\right) \\ 450=672\sin \left(2\mathrm{θ}\right) \\ \sin \left(2\mathrm{θ}\right)=\frac{450}{672} \\ \sin \left(2\mathrm{θ}\right)≈0.67 \\ 2\mathrm{θ}={\sin }^{-1}\left(0.67\right) \\ 2\mathrm{θ}=42 \\ \mathrm{θ}=21° \end{gathered}$$

By using the function of the ball we get,

$$\begin{gathered} R\left(\mathrm{θ}\right)=672\sin \left(2\mathrm{θ}\right) \\ 540=672\sin \left(2\mathrm{θ}\right) \\ \sin \left(2\mathrm{θ}\right)=\frac{540}{672} \\ \sin \left(2\mathrm{θ}\right)≈0.80 \\ 2\mathrm{θ}={\sin }^{-1}\left(0.80\right) \\ 2\mathrm{θ}=53.47 \\ \mathrm{θ}≈27° \end{gathered}$$

By using the function of the ball we get,

$$\begin{gathered} R\left(\mathrm{θ}\right)â‰\yen 480 \\ 672\sin \left(2\mathrm{θ}\right)â‰\yen 480 \\ \sin \left(2\mathrm{θ}\right)â‰\yen \frac{480}{672}≈0.71 \\ 2\mathrm{θ}â‰\yen {\sin }^{-1}\left(0.71\right) \\ 2\mathrm{θ}=45.5 \\ \mathrm{θ}≈23° \end{gathered}$$

To find that the golfer can hit the ball 720 feet we will use the function of a ball.

$$\begin{gathered} R\left(\mathrm{θ}\right)=672\sin \left(2\mathrm{θ}\right) \\ 720=672\sin \left(2\mathrm{θ}\right) \\ \sin \left(2\mathrm{θ}\right)=\frac{720}{672} \\ \sin \left(2\mathrm{θ}\right)≈1.07 \\ 2\mathrm{θ}={\sin }^{-1}\left(1.07\right)\left[{\sin }^{-1}\left(1.07\right)>90°\right] \\ 2\mathrm{θ}>90° \\ \mathrm{θ}>45° \end{gathered}$$

Thus, the golfer can hit the ball at$\mathrm{θ}>45°.$