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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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Verified Answer
The correct answer is:
$\tan ^{-1} \frac{b c}{a(c-a)}$
$\tan ^{-1} \frac{b c}{a(c-a)}$
$a=(u \cos \alpha) t$ 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, $c=\frac{u^2 \sin 2 \alpha}{g}$
$b=a \tan \alpha-\frac{a^2 g}{2}\left(\frac{\sin 2 \alpha}{c g}\right) \sec ^2 \alpha$
$b=a \tan \alpha-\frac{a^2}{2 c} 2 \tan \alpha$
$\Rightarrow\left(a-\frac{a^2}{c}\right) \tan \alpha=b$
$\tan \alpha=\frac{b c}{a(c-a)}$.
$b=a \tan \alpha-\frac{1}{2} g \frac{a^2}{u^2 \cos ^2 \alpha}$
also, $c=\frac{u^2 \sin 2 \alpha}{g}$
$b=a \tan \alpha-\frac{a^2 g}{2}\left(\frac{\sin 2 \alpha}{c g}\right) \sec ^2 \alpha$
$b=a \tan \alpha-\frac{a^2}{2 c} 2 \tan \alpha$
$\Rightarrow\left(a-\frac{a^2}{c}\right) \tan \alpha=b$
$\tan \alpha=\frac{b c}{a(c-a)}$.
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