A ball of mass 400, g is dropped from a heigh | vedclass

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(A) $1$. Velocity of the ball just before hitting the bat:$v_1 = \sqrt{2gh_1} = \sqrt{2 	imes 10 	imes 5} = 10\, m/s$ (downwards).$2$. Velocity of t

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$0.12$

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(A) $1$. Velocity of the ball just before hitting the bat:$v_1 = \sqrt{2gh_1} = \sqrt{2 	imes 10 	imes 5} = 10\, m/s$ (downwards).$2$. Velocity of the ball just after leaving the bat:$v_2 = \sqrt{2gh_2} = \sqrt{2 	imes 10 	imes 20} = 20\, m/s$ (upwards).$3$. Using the impulse-momentum theorem:$F_{avg} 	imes \Delta t = m(v_{final} - v_{initial})$Taking the upward direction as positive (+),we have $v_{final} = +20\, m/s$ and $v_{initial} = -10\, m/s$.Given $m = 400\, g = 0.4\, kg$ and $F_{avg} = 100\, N$.$100 	imes \Delta t = 0.4 	imes (20 - (-10))$$100 	imes \Delta t = 0.4 	imes 30$$100 	imes \Delta t = 12$$\Delta t = 0.12\, s$.

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