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 $$V=14\sqrt{\dfrac{R}{11}}$$.

Solution

## K.E of projection $$=P.E$$ at height $$10R$$$$\dfrac{1}{2}mv^2=mg10R$$$$v^2=20gR$$$$v=\sqrt{20gR}$$$$g=\dfrac{Gm}{R^2}$$$$V=\sqrt{20\times \dfrac{Gm}{R^2}\times R}$$$$V=\sqrt{20\dfrac{Gm}{R}}$$General Knowledge

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