Question

A body is thrown from the surface of the earth with velocity $\sqrt{\left(g{R}_e\right)/2}$, where $R_e$ is the radius of the earth at some angle from teh vertical. If the maximum height reached by the body is $R_e/4$, then the angle of projection with the vertical is

• A. ${\sin }^{-1}\left(\frac{\sqrt{5}}{4}\right)$
• B. ${\cos }^{-1}\left(\frac{\sqrt{5}}{4}\right)$
• C. ${\sin }^{-1}\left(\frac{\sqrt{3}}{4}\right)$
• D. none of these

Answer 1

The correct option is A ${\sin }^{-1}\left(\frac{\sqrt{5}}{4}\right)$

Conservation of angular momentum gives,

$m\left(u\sin \theta \right)R=mv\left(\frac{5}{4}R\right)$

Also from conservation of energy we get,

$\frac{1}{2}m{u}^2-\frac{GMm}{R}=\frac{1}{2}m{v}^2-\frac{GMm}{\frac{5}{4}R}$

It is given that $u=\sqrt{\frac{gR}{2}}$,

Using this value in above equations gives $\sin \theta =\frac{\sqrt{5}}{4}$

$⟹\theta =si{n}^{-1}\left(\frac{\sqrt{5}}{4}\right)$