A particle is projected with a velocity $v$ so that its range on a horizontal plane is twice the greatest height attained. Consider $g=10{\text{m/s}}^2$ as acceleration due to gravity. Then, its range will be:

The correct option is A $\frac{4v^2}{5g}$
Range is given by
$R=\frac{u^2\sin 2\theta }{g}$
and maximum height is given by
$H=\frac{u^2{\sin }^2\theta }{2g}$
and $R=2H$ [Given in question]
$\Rightarrow \frac{u^2\sin 2\theta }{g}=2\times \frac{u^2{\sin }^2\theta }{2g}$
$\Rightarrow 2\sin \theta \cos \theta ={\sin }^2\theta$
$\Rightarrow \tan \theta =2$


From triangle $ABC$,
$\sin \theta =\frac{2}{\sqrt{5}},\cos \theta =\frac{1}{\sqrt{5}}$

$\therefore R=\frac{2u^2\sin \theta \cos \theta }{g}=\frac{2v^2\times \frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}}{g}$
$\Rightarrow R=\frac{4v^2}{5g}$