A cannon ball has a range R on a horizontal plane. If h and h' are the greatest heights in the two paths for which this is possible, then

The range is same for angles $\alpha$ and $\left(\frac{\pi }{2}-\alpha \right)$.

The range R for $\alpha$ is given by

$\mathrm{R}=\frac{2{\mathrm{u}}^2\sin \mathrm{\alpha cos}\mathrm{\alpha }}{\mathrm{g}}$      ….(1)

Maximum height, $\mathrm{h}=\frac{{\mathrm{u}}^2{\sin }^2\mathrm{\alpha }}{2\mathrm{g}}$ … (2)

The range and height corresponding to angle $\left(\frac{\pi }{2}-\alpha \right)$ are

$\mathrm{R}=\frac{2{\mathrm{u}}^2\sin \left(\frac{\mathrm{\pi }}{2}-\mathrm{\alpha }\right)\cos \left(\frac{\mathrm{\pi }}{2}-\mathrm{\alpha }\right)}{\mathrm{g}}$

and   ${\mathrm{h}}^{\mathrm{\prime }}=\frac{{\mathrm{u}}^2{\sin }^2\left(\frac{\mathrm{\pi }}{2}-\mathrm{\alpha }\right)}{2\mathrm{g}}=\frac{{\mathrm{u}}^2{\cos }^2\mathrm{\alpha }}{2\mathrm{g}}$

Now   $\mathrm{h}\times {\mathrm{h}}^{\mathrm{\prime }}=\frac{{\mathrm{u}}^4{\sin }^2\mathrm{\alpha }\times {\cos }^2\mathrm{\alpha }}{4{\mathrm{g}}^2}$

$\begin{array}{l}\sqrt{{\mathrm{hh}}^{\mathrm{\prime }}}=\frac{{\mathrm{u}}^2\sin \mathrm{\alpha cos}\mathrm{\alpha }}{2\mathrm{g}}=\frac{\mathrm{R}}{4}\\ \therefore \mathrm{R}=4\sqrt{\left({\mathrm{hh}}^{\mathrm{\prime }}\right)}\end{array}$