Question

A particle is projected with a velocity $v$ such that its range on horizontal plane is twice the greatest height attained by it. The range of the projectile will be ($g$ is the 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

The correct option is A $\frac{4v^2}{5g}$

From the given condition of the projectile we have
$R=2H$
Where, $R$ is the range of the projectile and $H$ is the height of the projectile.

We know that, $R=4H\cot \theta \Rightarrow \cot \theta =\frac{2H}{4H}=\frac{1}{2}$
Plotting the value of $\cot \theta$ on triangle we get


$\sin \theta =\frac{2}{\sqrt{5}},\cos \theta =\frac{1}{\sqrt{5}}$
$\therefore$ Range of projectile is $R=\frac{2v^2\sin \theta \cos \theta }{g}$
$=\frac{2v^2}{g}\times \frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}=\frac{4v^2}{5g}.$