Q. A projectile is projected with velocity ${kv}_{e}$ in vertically upward direction from the ground into the space$\ (v_{e}$ is the escape velocity and$k < 1)$. If air resistance is considered to be negligible then the maximum height from the center of earth to which it can go will be $(R = radus\ of\ earth)$
A) $\frac{R}{k^{2} + 1}$
B) $\frac{R}{k^{2} - 1}$
C) $\frac{R}{1 - k^{2}}$
D) $\frac{R}{k + 1}$
Explanation
(c) $Kinetic\ energy = potential\ energy$
$\frac{1}{2}m\left( {kv}_{e} \right)^{2} = \frac{mgh}{1 + \frac{h}{R}}$
$\Longrightarrow \frac{1}{2}mk^{2}2gR = \frac{mgh}{1 + \frac{h}{R}} \Longrightarrow h = \frac{Rk^{2}}{1 - k^{2}}$
$Height\ of\ projectile\ from\ the\ earth’s\ surface = h$
$Height\ from\ centre\ r = R + h = R + \frac{Rk^{2}}{1 - k^{2}}$
$By\ solving\ r = \frac{R}{1 - k^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.