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

Answer 1

Answer: (A)

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

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