Question
If R be the horizontal range for inclination $\theta$ and h be the maximum height reached by the projectile, show that maximum range is given by $\frac{\text{R}^2}{8\text{h}}+2\text{h.}$
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Answer
Horizontal range of the projectile is, $\text{R}=\frac{\text{u}^2\sin2\theta}{\text{g}}$ Maximum height attained by the projectile is,$\text{h}=\frac{\text{u}^2\sin^2\theta}{2\text{g}}$
$\therefore\ \frac{\text{R}^2}{8\text{h}}+2\text{h}$
$=\frac{\text{u}^2(\sin2\theta)^2}{\text{g}^2}\times\frac{2\text{g}}{8\text{u}^2\sin^2\theta}+\frac{2\text{u}^2\sin^2\theta}{2\text{g}}$
$=\frac{\text{u}^2(2\sin\theta\cos\theta)^2}{4\text{g}\sin^2\theta}+\frac{\text{u}^2\sin^2\theta}{\text{g}}$
$=\frac{\text{u}^2\cos^2\theta}{\text{g}}+\frac{\text{u}^2\sin^2\theta}{\text{g}}$
$=\frac{\text{u}^2}{\text{g}}(\cos^2\theta+\sin^2\theta)=\frac{\text{u}^2}{\text{g}}=\text{R}_\text{max}$
$\therefore\ \frac{\text{R}^2}{8\text{h}}+2\text{h}$
$=\frac{\text{u}^2(\sin2\theta)^2}{\text{g}^2}\times\frac{2\text{g}}{8\text{u}^2\sin^2\theta}+\frac{2\text{u}^2\sin^2\theta}{2\text{g}}$
$=\frac{\text{u}^2(2\sin\theta\cos\theta)^2}{4\text{g}\sin^2\theta}+\frac{\text{u}^2\sin^2\theta}{\text{g}}$
$=\frac{\text{u}^2\cos^2\theta}{\text{g}}+\frac{\text{u}^2\sin^2\theta}{\text{g}}$
$=\frac{\text{u}^2}{\text{g}}(\cos^2\theta+\sin^2\theta)=\frac{\text{u}^2}{\text{g}}=\text{R}_\text{max}$
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