Question

A body is projected from the surface of the earth so that is reaches at a height of $10R.g=9.8m/s^2.$ show that the velocity with which the body should be projected from the surface of earth is $V=14\sqrt{\frac{R}{11}}$.

Answer 1

K.E of projection $=P.E$ at height $10R$

$\frac{1}{2}m{v}^2=mg10R$

$v^2=20gR$

$v=\sqrt{20gR}$

$g=\frac{Gm}{R^2}$

$V=\sqrt{20\times \dfrac

{Gm}{R^2}\times R}$

$V=\sqrt{20\frac{Gm}{R}}$