A projectile is fired from the surface of earth of radius `R` with a velocity `k upsilon_(e)` (where `upsilon_(e)` is the escape velocity from surface of earth and `k lt 1)`. Neglecting air resistance, the maximum height of rise from centre of earth is

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)=(GMm)/R=(GMm)/(R+h)=(GMmh)/(R(R+h))=(mghR)/(R+h)`
Since `v=nv_(3)` and `v_(3)=sqrt(2gR)` hence `=(Rn^(2))/(1+n^(2))`