Explanation

(b) Let a body be projected at a velocity $u$ at an angle $\theta$ with the horizontal. Then horizontal range covered is given by
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$R = \frac{u^{2}\sin{2\theta}}{g}\ \ldots(i)$
and height H is
$H = \frac{u^{2}\sin^{2}\theta}{2g}\ \ldots(ii)$
Given, $R = 3H$
$\frac{u^{2}\sin{2\theta}}{g} = 3 \times \frac{u^{2}\sin^{2}\theta}{2g}$
Also, sin2$\ \theta = 2\sin{{\theta cos}\theta}$
$\therefore\ \ \frac{u^{2}2\sin{\theta\cos\theta}}{g} = 3 \times \frac{u^{2}\sin^{2}\theta}{2g}$
$or\ \ 2\cos{\theta = 1.5\sin\theta\ }$
$or\tan{\theta = \frac{2}{1.5} = 1.33}$
$or\ \theta = 53{^\circ}7^{''}$
Hence, angle of projection is $53{^\circ}7''$