A ball of mass 0.5 kg is dropped from a heig | vedclass

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(A) The velocity of the ball just before hitting the ground is $v_i = -\sqrt{2gh_1} = -\sqrt{2 	imes 9.8 	imes 40} = -\sqrt{784} = -28 \ m/s$. The v

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$21 \ Ns$

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(A) The velocity of the ball just before hitting the ground is $v_i = -\sqrt{2gh_1} = -\sqrt{2 	imes 9.8 	imes 40} = -\sqrt{784} = -28 \ m/s$. The velocity of the ball just after rebounding is $v_f = \sqrt{2gh_2} = \sqrt{2 	imes 9.8 	imes 10} = \sqrt{196} = 14 \ m/s$. The impulse imparted to the ball is equal to the change in momentum: $\vec{I} = \Delta \vec{P} = m(\vec{v}_f - \vec{v}_i)$. Substituting the values: $\vec{I} = 0.5 	imes [14 - (-28)] = 0.5 	imes [14 + 28] = 0.5 	imes 42 = 21 \ Ns$.

$A$ ball of mass $0.5 \ kg$ is dropped from a height of $40 \ m$. The ball hits the ground and rises to a height of $10 \ m$. The impulse imparted to the ball during its collision with the ground is (Take $g = 9.8 \ m/s^2$)

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