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Physics Question on Gravitation

A particle of mass ‘m’ is projected with a velocity v=kVe(k < 1) from the surface of the earth. (Ve=escape velocity) The maximum height above the surface reached by the particle is

A

Rk21−k2\frac{Rk^2}{1-k^2}

B

R(k1−k)2R(\frac{k}{1-k})^2

C

R(k1+k)2R(\frac{k}{1+k})^2

D

R2k1+k\frac{R^2k}{1+k}

Answer

Rk21−k2\frac{Rk^2}{1-k^2}

Explanation

Solution

The correct answer is option (A) : Rk21−k2\frac{Rk^2}{1-k^2}