A particle of mass 'm' is projected with a velocity ` v = k V_e (k <1)` from the surface of the earth.(Ve = escape velocity)The maximum height above the surface reached by the particle is:

Options

A. `R((k)/(1+k))^2`
B. `(R^2k)/(1+k)`
C. `(Rk^2)/(1-k^2)`
D. `R((k)/(1-k))^2`

Answer

C

Explanation

`(-GmM)/R+1/2mv^2=-(GmM)/(R+h)`
`-(GmM)/(R+h)=(GmM)/R+1/2mk^2(2GM)/R`
`-(GmM)/(R+h)=(GmM)/R(k^2-1)`
`R+h=R/(1-k^2)`
`h=R[1/(1-k^2)-1]=R[(1-1+k^2)/(1-k^2)]rArrh=(Rk^2)/(1-k^2)`