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Q.
A particle is thrown vertically upwards with velocity $11.2\, km \,s^{-1}$ from the surface of earth. Calculate its velocity at height $3R$. Where $R$ is the radius of earth.
Let a particle of mass m is thrown vertically upwards with velocity v. Let its velocity be v' at a height 3R from the surface of the earth. According to the conservation of energy,
$\frac{1}{2}mv^{2}-\frac{GMm}{R}=\frac{1}{2}mv'^{2}-\frac{GMm}{\left(R+3R\right)}$
$v^{2}-\frac{2GM}{R}=v'^{2}-\frac{2GM}{4 R}$
$v^{2}-v'^{2}=\frac{2GM}{4R}+\frac{2GM}{R}=\frac{6}{4} \frac{GM}{R}$
$v'^{2}=v^{2}-\frac{6}{4} \frac{GM}{R}$
$v'^{2}=\sqrt{\left(11.2\right)^{2}-\frac{3}{2}gR}\quad\left(∵ g=\frac{GM}{R^{2}}\right)$
$\,=\sqrt{\left(11.2\right)^{2}-\frac{3}{2}\times\frac{9.8\times6400}{1000}}$
$\,=\sqrt{\left(11.2\right)^{2}-\left(94.08\right)}$
$\,=5.6 km s^{-1}$