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

Answer

B

Explanation

Using conservation of energy
`1/2m(Kv_(e))^(2)-(GMm)/R=-(GMm)/(R+h)`
Use `v_(e)=sqrt(2gR)` and `Gm=gR^(2)`
and to solve we get `h+R=R//(1=K^(2))`.