Question

A particle is projected with a velocity ' $v$ ' such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is ( $g$ = acceleration due to gravity)

• A. $\frac{4 v^{2}}{5 g}$
• B. $\frac{4 g}{5 v^{2}}$
• C. $\frac{v^{2}}{g}$
• D. $\frac{4 v^{2}}{\sqrt{5} g}$

Answer 1

Answer: (A)

Velocity of particle $=v$
If $\theta$ is angle of projection such a way,
$R=2 H$
$\Rightarrow \frac{v^{2} \sin 2 \theta}{g}=\frac{2 \cdot v^{2} \sin ^{2} \theta}{2 g}$
$2 \sin \theta \cos \theta=\sin ^{2} \theta$
$\Rightarrow 2=\frac{\sin \theta}{\cos \theta}$
$\Rightarrow 2=\tan \theta$
$\Rightarrow \tan \theta=2$

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