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 (where $g$ is accelaration due to gravity)

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

Answer 1

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

Find the value of $\sin \theta \&\cos \theta$ from the given relation.
Given: $R=2H\dots \left(i\right)$
As we know,
$R=4H\cot \theta \dots \left(ii\right)$
putting $\left(i\right)$ in $\left(ii\right)$ equation we get,
$\cot \theta =\frac{1}{2}$
Hence, $\tan \theta =2$
From traingle

diagrame

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

Find the rnge $\left(R\right)$ of projectile.

Now, the range of projectile is,
$R=\frac{v^2\sin 2\theta }{g}R=\frac{2v^2\sin \theta \cos \theta }{g}$
By putting the values, we get,
$R=\frac{2v^2}{g}\times \frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}R=\frac{4v^2}{5g}$