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. `R//((gR)/(2V^(2))-1)`
B. `R((gR)/(2V^(2))-1)`
C. `R//((2gR)/(V^(2))-1)`
D. `R((2gR)/(V^(2))-1)`

Answer

C

Explanation

`DeltaK.E=DeltaU`
`1/2MV^(2)=GM_(e)M(1/R-1/(R+h))......(i)`
Also `g=(GM_(e))/(R^(2))........(ii)`
On solving (i) and (ii) `h=R/(((2gR)/(V^(2))-1))`