Question

A body is thrown from the surface of earth with velocity $\sqrt{\frac{g{R}_e}{2}}$ at some angle from the vertical If the maximum height reached by the body is $\frac{R_e}{4}$ Then choose the correct option (S) $(R_e$ = Radius of eath)

• A. the angle of projection from the vertical is ${\sin }^{-1}\left(\frac{\sqrt{5}}{4}\right)$
• B. the velocity of the body at maximum height is $v=\sqrt{\frac{g{R}_e}{10}}$
• C. the angle of projection from the vertical is ${\sin }^{-1}\left(\frac{\sqrt{5}}{6}\right)$
• D. the velocity of the body at maximum height is zero

Answer 1

Answer: (A), (B)

The correct options are
A the angle of projection from the vertical is ${\sin }^{-1}\left(\frac{\sqrt{5}}{4}\right)$
B the velocity of the body at maximum height is $v=\sqrt{\frac{g{R}_e}{10}}$

From conservation of energy
$\frac{1}{2}m{\left(\sqrt{g{R}_e/2}\right)}^2-\frac{GMm}{R_e}=\frac{1}{2}m{v}^2-\frac{GMm}{(R_e+h)}$ .........(1)

From conservation of angular momentum
$m\sqrt{g{R}_e/2R_e}\sin \theta =mv(R_e+h)$.....(2)

Solving (1) & (2)
$\theta =\sin -1\left(\frac{\sqrt{5}}{4}\right)$ and $v=\sqrt{dfracg{R}_{e}10}$