A particle just clears a wall of height $b$ at distance $a$ and strikes the ground at a distance $c$ from the point of projection. The angle of projection is

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Aicharya · Accepted Answer

$a=(u \cos \alpha) t$ and $b=(u \sin \alpha) t-\frac{1}{2} g t^2$ $b=a \tan \alpha-\frac{1}{2} g \frac{a^2}{u^2 \cos ^2 \alpha}$ also, $c=\frac{u^2 \sin 2 \alpha}{g}$ $b=a \tan \alpha-\frac{a^2 g}{2}\left(\frac{\sin 2 \alpha}{c g}\right) \sec ^2 \alpha$ $b=a \tan \alpha-\frac{a^2}{2 c} 2 \tan \alpha$ $\Rightarrow\left(a-\frac{a^2}{c}\right) \tan \alpha=b$ $\tan \alpha=\frac{b c}{a(c-a)}$.

A particle just clears a wall of height $b$ at distance $a$ and strikes the ground at a distance $c$ from…

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