 Solution

For maximum projectile range from an elevated launch, let $X$ be horizontal projectile range from the tower and $\alpha$ the launch angle above horizontal. The projectile motion trajectory isPut $z=\tan \alpha$. For a fixed $X$, its largest possible $y$ occurs at $z=u^2/(gX)$; the envelope is $y_{\max }=h+u^2/(2g)-g{X}^2/(2u^2)$. The furthest attainable ground point satisfies $y_{\max }=0$, soThe optimum angle is less than $45^{\circ }$ for $h>0$ and approaches $45^{\circ }$ as $h\to 0$. The result is the maximum horizontal distance, using uniform gravity and no drag.