A ball of mass 50 ,g is dropped from a height | vedclass

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(C) $1$. Calculate the velocity of the ball just before hitting the bat $(v_1)$: Using $v^2 = u^2 + 2gh$,where $u = 0$,$h = 20 \,m$,and $g = 10 \,m/s^

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$1/80^{	ext{th}}$

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(C) $1$. Calculate the velocity of the ball just before hitting the bat $(v_1)$: Using $v^2 = u^2 + 2gh$,where $u = 0$,$h = 20 \,m$,and $g = 10 \,m/s^2$,we get $v_1 = \sqrt{2 	imes 10 	imes 20} = \sqrt{400} = 20 \,m/s$ (downwards).$2$. Calculate the velocity of the ball just after being hit by the bat $(v_2)$: Using $v^2 = u^2 + 2gh$ for the upward motion,where $v = 0$ at the maximum height $h = 45 \,m$,we get $0 = u^2 - 2 	imes 10 	imes 45$,so $u = v_2 = \sqrt{900} = 30 \,m/s$ (upwards).$3$. Apply the impulse-momentum theorem: $	ext{Impulse} = F \Delta t = m(v_{	ext{final}} - v_{	ext{initial}})$.$4$. Taking upward direction as positive: $v_{	ext{final}} = +30 \,m/s$ and $v_{	ext{initial}} = -20 \,m/s$.$5$. The force exerted by the bat is upward,so $F = +200 \,N$. The weight of the ball $(mg = 0.05 	imes 10 = 0.5 \,N)$ acts downwards. The net average force is $F_{	ext{net}} = F_{	ext{bat}} - mg = 200 - 0.5 \approx 200 \,N$.$6$. $\Delta t = \frac{m(v_2 - v_1)}{F_{	ext{net}}} = \frac{0.05 	imes (30 - (-20))}{200} = \frac{0.05 	imes 50}{200} = \frac{2.5}{200} = \frac{1}{80} \,s$.

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