A stone thrown with velocity 'u' at angles '$\theta$' and (90°-$\theta$) with the horizontal reaches to maximum heights H₁ and H₂ respectively. Its horizontal range is

We know that, 
${\mathrm{H}}_1=\frac{{\mathrm{u}}^2{\sin }^2\theta }{2\mathrm{g}}$ and ${\mathrm{H}}_2=\frac{{\mathrm{u}}^2{\sin }^2(90-\theta )}{2\mathrm{g}}$

$\begin{array}{l}\Rightarrow {\mathrm{H}}_2=\frac{{\mathrm{u}}^2{\cos }^2\theta }{2\mathrm{g}}\\ \because {\mathrm{H}}_1{\mathrm{H}}_2=\frac{{\mathrm{u}}^2{\sin }^2\theta }{2\mathrm{g}}\times \frac{{\mathrm{u}}^2{\cos }^2\theta }{2\mathrm{g}}\\ =\frac{4}{4}\times \frac{{\mathrm{U}}^2{\sin }^2\theta }{2\mathrm{g}}\times \frac{{\mathrm{U}}^2{\cos }^2\theta }{2\mathrm{g}}\\ =\frac{{\left({\mathrm{U}}^2\sin 2\theta \right)}^2}{16{\mathrm{g}}^2}\left[\because \sin 2\theta =2\sin \theta \cos \theta \right]\\ =\frac{{\mathrm{R}}^2}{16}\\ {\mathrm{R}}^2=6{\mathrm{H}}_1{\mathrm{H}}_2\\ \therefore \mathrm{R}=4\sqrt{{\mathrm{H}}_1{\mathrm{H}}_2}\end{array}$