CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image