91Ó°ÊÓ

Given the object is propelled at an angle$45°<\mathrm{θ}<90°$

Calculating, we get

$$\begin{gathered} \sin \mathrm{θ}\cos \mathrm{θ}=\frac{\sin \left(2\mathrm{θ}\right)}{2} \\ {\cos }^2\mathrm{θ}=\frac{\cos \left(2\mathrm{θ}\right)+1}{2} \end{gathered}$$

Calculating, we get

$$\begin{gathered} R\left(\mathrm{θ}\right)=\frac{v_0^2\sqrt{2}}{16}\cos \mathrm{θ}(\sin \mathrm{θ}-\cos \mathrm{θ}) \\ R\left(\mathrm{θ}\right)=\frac{v_0^2\sqrt{2}}{16}\cos \mathrm{θ}(\sin \mathrm{θ}-\cos \mathrm{θ}) \\ R\left(\mathrm{θ}\right)=\frac{v_0^2\sqrt{2}}{16}\left(\sin \mathrm{θ}\cos \mathrm{θ}-{\cos }^2\mathrm{θ}\right) \\ R\left(\mathrm{θ}\right)=\frac{v_0^2\sqrt{2}}{16}\left(\sin \mathrm{θ}\cos \mathrm{θ}-{\cos }^2\mathrm{θ}\right) \\ R\left(\mathrm{θ}\right)=\frac{v_0^2\sqrt{2}}{16}\left[\frac{\sin \left(2\mathrm{θ}\right)}{2}-{\cos }^2\mathrm{θ}\right] \\ R\left(\mathrm{θ}\right)=\frac{v_0^2\sqrt{2}}{16}\left[\frac{\sin \left(2\mathrm{θ}\right)}{2}-\frac{\cos \left(2\mathrm{θ}\right)+1}{2}\right] \\ R\left(\mathrm{θ}\right)=\frac{v_0^2\sqrt{2}}{16}\left(\frac{1}{2}\right)\left[\sin \right(2\mathrm{θ})-\cos (2\mathrm{θ})-1] \\ R\left(\mathrm{θ}\right)=\frac{v_0^2\sqrt{2}}{32}\left[\sin \right(2\mathrm{θ})-\cos (2\mathrm{θ})-1] \end{gathered}$$

Given the expression

Calculating, we get

$$\begin{gathered} \frac{\sin \left(2\mathrm{θ}\right)}{\cos \left(2\mathrm{θ}\right)}+1=0 \\ \tan 2\mathrm{θ}+1=0 \\ \tan 2\mathrm{θ}=-1 \\ 2\mathrm{θ}=\arctan (-1) \\ 2\mathrm{θ}=135°,315° \\ \mathrm{θ}=67.5°,157.5° \end{gathered}$$

Since$45°<\mathrm{θ}<90°,\mathrm{θ}=67.5°$

Given$v_0=32ft/sec$

Substituting, we get

$$\begin{gathered} R\left(\mathrm{θ}\right)=\frac{v_0^2\sqrt{2}}{32}\left[\sin \right(2\mathrm{θ})-\cos (2\mathrm{θ})-1] \\ R\left(67.5°\right)=\frac{(32{)}^2\sqrt{2}}{32}\left[\sin \left(2·67.5°\right)-\cos \left(2·67.5°\right)-1\right] \\ R\left(67.5°\right)=32\sqrt{2}\left[\sin \left(135°\right)-\cos \left(135°\right)-1\right] \\ R\left(67.5°\right)=32\sqrt{2}\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-1\right) \\ R\left(67.5°\right)=32\sqrt{2}(\sqrt{2}-1) \\ R\left(67.5°\right)=64-32\sqrt{2} \end{gathered}$$

Given$R=R\left(\mathrm{θ}\right),45°≤\mathrm{θ}≤90°$

The graph is

From the graph, we can see that the maximum distance $R$is achieved at $\mathrm{θ}=67.5°$with value $(64-32\sqrt{2})\mathrm{ft}$, identical to the answers from the computations above.