Q. A projectile is projected with velocity $kv_{e}$ in vertically upward direction from the ground into the space. ($v_{e}$ is escape velocity and $k < 1$). If resistance is considered to be negligible then the maximum height from the centre of earth to which it can go, will be $:(R =$ radius 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
$\frac{1}{2}m\left( kv_{e} \right)^{2} = \frac{mgh}{1 + \frac{h}{R}} \Rightarrow \frac{1}{2}mk^{2}2gR = \frac{mgh}{1 + \frac{h}{R}} \Rightarrow h = \frac{Rk^{2}}{1 - k^{2}}$
Height of Projectile from the earth’s surface $= h$
Height from the 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.