Answer
An angle of $41^{\circ}$ will maximize the distance.
Work Step by Step
We can find the angle $\theta$ which will maximize the distance:
$\theta = arcsin \left(\sqrt{\frac{v^2}{2v^2+64h}}\right)$
$\theta = arcsin \left(\sqrt{\frac{(32)^2}{(2)(32)^2+(64)(5.0)}}\right)$
$\theta = arcsin \left(\sqrt{\frac{1024}{2048+320}}\right)$
$\theta = arcsin \left(\sqrt{\frac{1024}{2368}}\right)$
$\theta = arcsin \left(\sqrt{0.432432}\right)$
$\theta = arcsin \left(0.657596\right)$
$\theta = 41^{\circ}$
An angle of $41^{\circ}$ will maximize the distance.
