Question

What is the angle $\theta$ of projection with horizontal plane of a projectile if its range is $\frac{\sqrt{3v^3}}{2g}$, where v is velocity of projection?

Answer 1

Given that,

Range $R=\frac{\sqrt{3v^3}}{2g}$

We know that,

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

Now, the angle of projection

$\frac{v^2\sin 2\theta }{g}=\frac{\sqrt{3v^3}}{2g}$

$\frac{v^4{\sin }^{2}2\theta }{g^2}=\frac{3v^3}{4g^2}$

${\sin }^{2}2\theta =\frac{3}{4v}$

$\sin 2\theta =\frac{\sqrt{3}}{2\sqrt{v}}$

$2\theta ={\sin }^{-1}\frac{\sqrt{3}}{2\sqrt{v}}$

$\theta ={\sin }^{-1}\frac{\sqrt{3}}{4\sqrt{v}}$

Hence, this is the required solution