A ball of mass m is fired vertically upwards from the surface of the earth with velocity `nv_​e`​, where `v_​e` is the escape velocity and n < 1. Neglecting air resistance, to what height will the ball rise ? (Take radius of the earth as R) :-

Options

A. `R//n^2`
B. `R//(1-n^2)`
C. `Rn^2//(1-n^2)`
D. `Rn^2`

Answer

C

Explanation

`1/2mv^2 = (GMn)/R -(GMn)/(R+h)`
`=(GMmh)/(R(R+h))`
`therefore 1/2 mv^2 =(mghR)/(R+h)`
Since , `v=nv_e` and `v_e=sqrt(2gR)` ,
hence `h=(Rn^2)/((1-n^2))`