Question
Prove that the maximum horizontal range is four times the maximum height attained by a projectile which is fired along the required oblique direction.

Answer

The required angle of projection for max. horizontal range is 45°. $\text{R}_\text{max}=\frac{\text{u}^2}{\text{g}}$ Max. height, $\text{H}=\frac{\text{u}^2\sin^2\theta}{2\text{g}}$ $=\frac{\text{u}^2\sin^245^\circ}{2\text{g}}=\frac{\text{u}^2}{4\text{g}}$ $\frac{\text{R}_\text{max}}{\text{H}}=\frac{\frac{\text{u}^2}{\text{g}}}{\frac{\text{u}^2}{4\text{g}}}=4$ or $\text{R}_\text{max}=4\text{H}$

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