Let the projectile be fired with velocity $u$ by making an angle $\theta$ with the horizontal. The height $h$ gained by the projectile is given by, \[h=\frac{u^2\sin^2\theta}{2g}\]

The maximum height is attained at $\theta=45°.$ \[\therefore H_{\text{max}}=\frac{u^2\sin^245°}{2g}=\frac{u^2}{4g}\]

The horizontal range $R$ is given by, \[R=\frac{u^2\sin2\theta}{g}\]

The maximum horizontal range is attained at $\theta=45°.$ \[\therefore R_{\text{max}}=\frac{u^2\sin^2(90°)}{g}=\frac{u^2}{g}=4×\frac{u^2}{4g}\] \[\therefore R_{\text{max}} = 4H_{\text{max}}\]

[**Read:** Projectile Motion]