A projectile is projectile with velocity `kv_(e)` in vertically upward direction from the ground into the space (`v_(e)` is escape velocity and `klt1`). If air resistance is considered to be negligible then the maximum height from the centre of earth to which it can go, will be : (`R`=raduis of earth)

Options

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

Answer

C

Explanation

Kinetic energy=Potential energy
`(1)/(2)m(kv_(e))^(2)=(mgh)/(1+h//R)`
`implies(1)/(2)mk^(2)2gR=(mgh)/((1+h//R))" "(becausev_(e)=sqrt(2gR))`
`impliesh=(Rk^(2))/(1-k^(2))`
Height of projectile from earth's surface=h
Height from the centre,
`r=R+h=R+(Rk^(2))/(1-k^(2))`
On solving, we get `r=(R)/(1-k^(2))`