From the pole of the earth, a body of mass m is imparted a velocity `v_0` directed vertically up. If Mis the mass of the earth, R its radius and g is the free-fall acceleration on its surface, then the height It to which the body will ascent is (neglect air resistance)

Options

A. `(R v_0^2)/((2gR - v_0^2))`
B. `(Rv_0^2)/(2gR)`
C. `R`
D. `(Rv_0^2)/((2gR + v_0^2) )`

Answer

A

Explanation

If body moves up to height h, then by conservation of energy , we get
` Delta (KE) - Delta (PE)`
` 0 -1/2 mv_0^2 = (- GMm)/(R ) - ( - (GMm)/(R + h) )`
` rArr (v_0^2)/(2Gm) = 1/R - (1)/(R + h) rArr (R (R + h))/(h) = (2gR^2)/(v_0^2)`
` rArr h = (R)/( (2gR)/(v_0^2) - 1) )" or " h = (Rv_0^2)/(2g R - v_0^2)`