The acceleration due to gravity on the surface of earth is g \& radius R . A ball is projected horizontally from point P 1 on the surface of earth (as shown in the figure) If the maximum height atta

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The acceleration due to gravity on the surface of earth is g \& radius R . A ball is projected horizontally from point P 1 ​ on the surface of earth (as shown in the figure) If the maximum height attained by the ball above the earth surface is 3 R then speed of projection V 0 ​ is (neglect cosmic dust resistance \& earth rotation about its axis)

TestHub · Accepted Answer

Let speed at max height is v Since, it is at max height, v is perpendicular to r ⇒ By conservation of angular momentum

$mv_{0} R = mv (4 R)$

$v = \frac{V _{0}}{4}$ ⇒ By conservation of mechanical energy $\frac{1}{2} m v_{0}^{2} - \frac{GM m}{R} = \frac{1}{2} m (\frac{v _{0}}{4})^{2} - \frac{GM m}{4 R}$ $v_{0} = \frac{8}{5} \frac{GM}{R}$ $g = \frac{GM}{R ^{2}}, v_{0} = \frac{8}{5} gR$

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