Answer
As $\theta$ increases from $40^{\circ}$ to $42^{\circ}$, the distance increases slightly. As $\theta$ increases from $42^{\circ}$ to $45^{\circ}$, the distance decreases slightly. It seems that $42^{\circ}$ is close to the optimal angle. A change in the angle seems to have a small effect on the distance.
Work Step by Step
$D = \frac{v^2~sin~\theta~cos~\theta+v~cos~\theta~\sqrt{(v~sin~\theta)^2+64~h}}{32}$
We can find $D$ when $\theta = 40^{\circ}$:
$D = \frac{(44~ft/s)^2~sin~40^{\circ}~cos~40^{\circ}+44~ft/s~cos~40^{\circ}~\sqrt{(44~ft/s~sin~40^{\circ})^2+(64)(7~ft)}}{32}$
$D = 67.0~ft$
We can find $D$ when $\theta = 42^{\circ}$:
$D = \frac{(44~ft/s)^2~sin~42^{\circ}~cos~42^{\circ}+44~ft/s~cos~42^{\circ}~\sqrt{(44~ft/s~sin~42^{\circ})^2+(64)(7~ft)}}{32}$
$D = 67.1~ft$
We can find $D$ when $\theta = 45^{\circ}$:
$D = \frac{(44~ft/s)^2~sin~45^{\circ}~cos~45^{\circ}+44~ft/s~cos~45^{\circ}~\sqrt{(44~ft/s~sin~45^{\circ})^2+(64)(7~ft)}}{32}$
$D = 66.8~ft$
As $\theta$ increases from $40^{\circ}$ to $42^{\circ}$, the distance increases slightly. As $\theta$ increases from $42^{\circ}$ to $45^{\circ}$, the distance decreases slightly. It seems that $42^{\circ}$ is close to the optimal angle. A change in the angle seems to have a small effect on the distance.
