Question

A particle is projected vertically upwards from the surface of the earth(radius R) with a kinetic energy equal to half of the minimum value needed for it to escape. Find the height to which it rises above the surface of earth.

Answer 1

$\frac{1}{2}\frac{MmG}{R}=m.\frac{GMR}{{(R+h)}^2}\Rightarrow {(R+h)}^2=2R^2\Rightarrow R+h=\sqrt{2}R\Rightarrow h=(\sqrt{2}-1)RE\times m=\frac{1}{2}{mv}^{2}v=\sqrt{2E}$