A ball of mass 200,g rests on a vertical post | vedclass

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(D) First,calculate the time taken by both objects to reach the ground from height $h = 20\,m$ using $h = \frac{1}{2}gt^2$:$t = \sqrt{\frac{2h}{g}} =

Vedclass · Accepted Answer

$360$

Vedclass · Answer

(D) First,calculate the time taken by both objects to reach the ground from height $h = 20\,m$ using $h = \frac{1}{2}gt^2$:$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 	imes 20}{10}} = \sqrt{4} = 2\,s$.Let $v_1$ be the velocity of the ball and $v_2$ be the velocity of the bullet after the collision.For the ball: $v_1 = \frac{	ext{distance}}{	ext{time}} = \frac{30}{2} = 15\,m/s$.For the bullet: $v_2 = \frac{	ext{distance}}{	ext{time}} = \frac{120}{2} = 60\,m/s$.Using the law of conservation of linear momentum in the horizontal direction:$m_{bullet} u = m_{ball} v_1 + m_{bullet} v_2$$(0.01) u = (0.2)(15) + (0.01)(60)$$0.01 u = 3 + 0.6 = 3.6$$u = \frac{3.6}{0.01} = 360\,m/s$.

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