A particle is projected under gravity with velocity $\sqrt{2\mathrm{ag}}$ from a point at a height h above the level plane at an angle $\mathrm{\theta }$ to it. The maximum range R on the ground is:

Coordinate of point P are (R, - h)

$-\mathrm{h} = \mathrm{R} \tan \mathrm{\theta }- \frac{{\mathrm{gR}}^2}{2\left(2\mathrm{ga}\right)}(1+{\tan }^2\mathrm{\theta })$

or, ${\mathrm{R}}^2{\tan }^2\mathrm{\theta }-4\mathrm{aR} \tan \mathrm{\theta }+({\mathrm{R}}^2-4\mathrm{ah}) = 0$

For $\mathrm{\theta }$ to be real,

${\left(4\mathrm{aR}\right)}^2 \ge 4{\mathrm{R}}^2({\mathrm{R}}^2-4\mathrm{ah})$

or, $4{\mathrm{a}}^2 \ge ({\mathrm{R}}^2-4\mathrm{ah})$

or, ${\mathrm{R}}^2 \le 4\mathrm{a}(\mathrm{a}+\mathrm{h})$

or, ${\mathrm{R}}^2 \le 2\sqrt{\mathrm{a}(\mathrm{a}+\mathrm{h})}$

Hence, $\mathrm{R} \le 2\sqrt{\mathrm{a}(\mathrm{a}+\mathrm{h})}$

     ${\mathrm{R}}_{\max } = 2\sqrt{\mathrm{a}(\mathrm{a}+\mathrm{h})}$