Explanation

(c) The range of particle is
$R = \frac{u^{2}sin2\theta}{g}$
$R = \frac{u^{2}sin2 \times 45{^\circ}}{g}\ (\therefore\theta = 45{^\circ})$
$or\ R = \frac{u^{2}sin90{^\circ}}{g}$
$or\ \ R = \frac{u^{2}}{g}\ \ \ \ldots.(i)$
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Now, the maximum height of the particle is
$H = \frac{u^{2}\sin^{2}\theta}{2g}$
$= \frac{u^{2}\sin^{2}45{^\circ}}{2g} = \frac{u^{2}\left( \frac{1}{\sqrt{2}} \right)^{2}}{2g}$
$= \frac{u^{2}}{4g}\ \ldots(ii)$
Dividing Eqs. (i) by Eq. (ii),
$\frac{R}{H} = \frac{u^{2}/g}{u^{2}/4g}$
$or\ \ \ R = 4H$