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

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(B) Given,the height of both towers is the same,$h_1 = h_2 = h = 20 \ m$.The time of flight $t$ for a body thrown horizontally is given by $t = \sqrt{

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(B) Given,the height of both towers is the same,$h_1 = h_2 = h = 20 \ m$.The time of flight $t$ for a body thrown horizontally is given by $t = \sqrt{\frac{2h}{g}}$.$t = \sqrt{\frac{2 	imes 20}{10}} = \sqrt{4} = 2 \ s$.The horizontal displacement of the body from tower $A$ to point $P$ is $x_A = u_A t = 20 \ ms^{-1} 	imes 2 \ s = 40 \ m$.The horizontal displacement of the body from tower $B$ to point $Q$ is $x_B = u_B t = 30 \ ms^{-1} 	imes 2 \ s = 60 \ m$.The distance between points $P$ and $Q$ is $d = 200 \ m - (x_A + x_B) = 200 \ m - (40 \ m + 60 \ m) = 100 \ m$.For the car starting from rest $(u = 0)$ at $P$ and reaching $Q$ in $t' = 10 \ s$,we use the equation of motion $s = ut + \frac{1}{2}at^2$:$100 = 0 	imes 10 + \frac{1}{2} 	imes a 	imes (10)^2$.$100 = 50a$.$a = 2 \ ms^{-2}$.

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