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 acceleration due to gravity)

The correct option is 4v²/5g.

Let us derive the range of the projectile step-by-step based on the given information.

Let the angle of projection with the horizontal be $\theta$, and the velocity of projection be $v$.
The formula for the horizontal range ($R$) of a projectile is given by:

$R=\frac{v^{2}sin\left(2\theta \right)}{g}=\frac{2v^{2}sin\theta cos\theta }{g}$

The formula for the greatest height ($H$) attained by the projectile is:

$H=\frac{v^2{sin}^2\theta }{2g}$

According to the problem, the horizontal range is twice the greatest height attained by the particle:

$R=2H$

Substituting the expressions for $R$ and $H$ into this relation, we get:

$\frac{2v^{2}sin\theta cos\theta }{g}=2\left(\frac{v^2{sin}^2\theta }{2g}\right)$

Simplifying both sides by canceling common terms ($v^2$, $g$, and $sin\theta$, assuming $sin\theta \ne 0$):

$2cos\theta =sin\theta$

This gives:

$tan\theta =2$

From $tan\theta =2$, we can determine $sin\theta$ and $cos\theta$ using a right-angled triangle where the opposite side is 2 and the adjacent side is 1. The hypotenuse is:

$\sqrt{2^2+1^2}=\sqrt{5}$

Therefore, we have:

$sin\theta =\frac{2}{\sqrt{5}}$

And:

$cos\theta =\frac{1}{\sqrt{5}}$

Now, substitute these values back into the expression for the horizontal range $R$:

$R=\frac{2v^{2}sin\theta cos\theta }{g}$

$R=\frac{2v^2}{g}·\left(\frac{2}{\sqrt{5}}\right)·\left(\frac{1}{\sqrt{5}}\right)$

$R=\frac{4v^2}{5g}$

Thus, the horizontal range of the projectile is indeed $\frac{4v^2}{5g}$.