A rocket is launched vertically from the surface of earth with an initial velocity `v`. How far above the surface of earth it will go? Neglect the air resistance.

Options

A. `sqrt(2gR)`
B. `u^(2)//2g`
C. `[(2g)/(u^(2)) - (1)/(R )]`
D. `[(2g)/(u^(2)) - (1)/(R )]^(-1)`

Answer

D

Explanation

Total energy of rocket on the surface of earth
`= (1)/(2)mu^(2) =- (GM m)/(R ) = (1)/(2) mu^(2) -mg R`
total energy of rocket at the height `h`
`=0 - (GM m)/((R + h)) = - (mg R^(2))/((R + h))`
According to law of conservation of energy
`(1)/(2)mu^(2) -mg R = -(mg R^(2))/((R + h))`
`(1)/(2) mu^(2) = mg R^(2)[(1)/(R ) -(1)/(R + h)] = (mg Rh)/((R + h))`
or `(R + h)/(h) = (2g R)/(u^(2))` or `(R )/(h) + 1 = (2g R)/(u^(2))`
or `(R )/(h) = (2g r)/(u^(2)) -1`
or `h = (R )/(((2g R)/(u^(2))-1)) = [(2g)/(u^(2)) -(1)/(R )]^(-1)`