Motion in a Plane

To obtain maximum range, angle of projection must be $45^{\circ}$, i.e. $\theta=45^{\circ}$.

$\therefore H_{\max }=\frac{u^2 \sin ^2 45^{\circ}}{2 g}=\frac{u^2}{2 g}\left(\frac{1}{\sqrt{2}}\right)^2=\frac{u^2}{4 g}=\frac{R_{\max }}{4}$

So, $H_{\max }$ is $25 \%$ of $R_{\max }$.

Horizontal range $=\frac{u^2 \sin 2 \theta}{g}$

When $\theta=45^{\circ}$, maximum horizontal range

$R_{\max }=\frac{u^2}{g} \sin 90^{\circ}=\frac{u^2}{g}$

When $\theta=135$, maximum horizontal range

$R=\frac{u^2}{g} \sin 270=\frac{-u^2}{g}$

Negative sign implies opposite direction.