Answer
$D = \frac{v^2~sin~(2~\theta)}{32}$
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 $h=0$:
$D = \frac{v^2~sin~\theta~cos~\theta+v~cos~\theta~\sqrt{(v~sin~\theta)^2+(64)(0)}}{32}$
$D = \frac{v^2~sin~\theta~cos~\theta+v~cos~\theta~\sqrt{(v~sin~\theta)^2}}{32}$
$D = \frac{v^2~sin~\theta~cos~\theta+v^2~sin~\theta~cos~\theta}{32}$
$D = \frac{2~v^2~sin~\theta~cos~\theta}{32}$
$D = \frac{v^2~sin~(2~\theta)}{32}$
