Two towers A and B,each of height 20 m,are s | vedclass

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(B) The time $t$ taken by both bodies to reach the ground is the same because the initial vertical velocity is zero and the height $h$ of both towers

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$2$

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(B) The time $t$ taken by both bodies to reach the ground is the same because the initial vertical velocity is zero and the height $h$ of both towers is the same.Using the equation $h = \frac{1}{2}gt^2$,we get $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 	imes 20}{10}} = 2 \ s$.The horizontal distance covered by the body from tower $A$ is $AP = u_A 	imes t = 20 \ ms^{-1} 	imes 2 \ s = 40 \ m$.The horizontal distance covered by the body from tower $B$ is $QB = u_B 	imes t = 30 \ ms^{-1} 	imes 2 \ s = 60 \ m$.The distance between $P$ and $Q$ is $PQ = 200 \ m - (AP + QB) = 200 \ m - (40 \ m + 60 \ m) = 100 \ m$.For the car,initial velocity $u = 0$,distance $s = 100 \ m$,and time $t = 10 \ s$.Using the equation $s = ut + \frac{1}{2}at^2$,we have $100 = 0 + \frac{1}{2} 	imes a 	imes (10)^2$.$100 = 50a$,which gives $a = 2 \ ms^{-2}$.

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