Question
Shows that the projection angle $\theta_0$ for a projectile launched from the origin is given by, $\theta_0=\tan^{-1}\Big(\frac{4\text{h}_\text{m}}{\text{R}}\Big)$ where the symbols have their usual meaning.
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Answer
Maximum vertical height, $\text{h}_\text{m}=\frac{\text{u}_0^2\sin^2\theta}{2\text{g}}\ ...(\text{i})$ Horizontal range, $\text{R}=\frac{\text{u}_0^2\sin^22\theta}{\text{g}}\ ...(\text{ii})$ $\frac{\text{h}_\text{m}}{\text{R}}=\frac{\sin^2\theta}{2\sin^22\theta}$ $=\frac{\sin\theta\times\sin\theta}{2\times\sin\theta\cos\theta}$ $=\frac{\sin\theta}{4\cos\theta}=\frac{\tan\theta}{4}$ $\tan\theta=\frac{4\text{h}_\text{m}}{\text{R}}$ $\theta=\tan^{-1}\frac{4\text{h}_\text{m}}{\text{R}}$
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