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

Explanation

Let a body of mass `m` be projected from the surface of Earth with speed `upsilon` and it reaches to a height `h`. Using law of conservation of energy (relative to surface of Earth) we have,
`- (GMm)/(R ) + (1)/(2) m upsilon^(2) = - (GMm)/((R + h))`
or` (1)/(2) m upsilon^(2) = (GMm)/(R ) - (GMm)/((R + h)) = (mgh)/((1 + h//R))`
In this problem, `upsilon = K upsilon_(e) K sqrt(2gR)`
and `h = r - R`
So, `(1)/(2) m K^(2) 2 gR = (mg (r - R))/([1 + (r - R)//R]`
or `K^(2) = (r - R)/(r ) = 1 - (R )/(r )` or `r = (R )/(1 - K^(2))`