 Maximum projectile range from an elevated launch

Under uniform gravity without drag, a projectile launched at speed $u$ from height $h$ reaches a maximum horizontal projectile range $u\sqrt{u^2+2gh}/g$. For horizontal position $X$, maximize $y=h+Xz-g{X}^2(1+z^2)/(2u^2)$ over $z=\tan \alpha$. The maximizer is $z=u^2/(gX)$ and its envelope meets the ground at the displayed distance. The optimum angle obeys $\tan \alpha =u/\sqrt{u^2+2gh}$.