Question

A particle is projected vertically up with velocity $v=\sqrt{\frac{4g{R}_e}{3}}$ from earth surface. The velocity of particle at height equal to half of the maximum height reached by it

• A. $\sqrt{\frac{g{R}_e}{2}}$
• B. $\sqrt{\frac{g{R}_e}{3}}$
• C. $\sqrt{g{R}_e}$
• D. $\sqrt{\frac{2g{R}_e}{3}}$

Answer 1

Answer: (B)

Conserving mechanical energy at the surface of earth and the maximum height attained,
$\frac{-GMm}{R_e}+\frac{1}{2}m\frac{4GM}{3R_e^2}{R}_e=\frac{-GMm}{r}+0$
$P.E_i+K.E_i=P.E_f+K.E_l$
$\Rightarrow \frac{-GMm}{R_a}+\frac{2GMm}{3R^e}=\frac{-GMm}{r}$
$=\frac{1}{3}\frac{GMm}{R_e}=\frac{-GMm}{r}$
$\Rightarrow r=3R_e$
$\Rightarrow R_e+h=3R_e$
$h=2R_e$
Now, let us calculate the velocity of the particle at height equal to half of the maximum height i.e at $h=R_e$ Again using mechanical conservation of energy,
$P.E_i+K.E_i=P.E_j+K\cdot E_j$
$\frac{-GMm}{R_e}+\frac{1}{2}m\frac{4}{3}\frac{GM}{R_e^2}\times R_e=\frac{-GMm}{2R}+\frac{1}{2}m{v}^2$
$\Rightarrow -\frac{1}{3}\frac{GMm}{R_e}+\frac{GMm}{2R_e}=\frac{1}{2}m{v}^2$
$\Rightarrow \frac{GMm}{6R_e}=\frac{1}{2}m{v}^2$
$\Rightarrow v=\sqrt{\frac{GM}{3R_e}}=\sqrt{\frac{GM}{R_e^2}\times \frac{R_e}{3}}=\sqrt{\frac{g{R}_e}{3}}$