A body which is initially at rest at a height R above the surface of the earth of radius R, falls freely towards the earth. Find out its velocity on reaching the surface of earth. Take `g=` acceleration due to gravity on the surface of the Earth.

Options

A. `sqrt((2gR))`
B. `sqrt((gR))`
C. `sqrt((3)/(2)gR)`
D. `sqrt((4gR))`

Answer

B

Explanation

Increase in kinetic energy = decrease in potential energy
`:. (1)/(2)mv^(2)=(mgR)/(1+R//R)=(mgR)/(2)" "(because DeltaU=(mgh)/(1+h//R))`
`rArr (1)/(2)mv^(2)=(mgR)/(2)rArr v=sqrt(gR)`.