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

Answer

B

Explanation

If a body is projected from the surface of the earth with a velocity v and reaches a height h, then according to law of conservation of energy
`1/2mv^(2)=(mgh)/(1+h//R)`
Here `v=kv_(e)=ksqrt(2gR)`
Therefore, `1/2mk^(23)gR=(mg(r-R))/(1+(r-R)/R)`
`k^(2)R[1+(r-R)/R]=r-R` or `k^(2)r=r-R`
or `r=(R/(1-k^(2)))`