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 $s = ut + \frac{1}{2}at^2$:$100 = 0 	imes 10 + \frac{1}{2} 	imes a 	imes (10)^2$.$100 = 50a$.$a = 2 \ ms^{-2}$.

Column-$I$ $(R/H_{max})$	Column-$II$ (Angle of projection $\theta$)
$A. 1$	$1. 60^o$
$B. 4$	$2. 30^o$
$C. 4\sqrt{3}$	$3. 45^o$
$D. 4/\sqrt{3}$	$4. \tan^{-1}(4) = 76^o$

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Match the columns:

Column-$I$ $(R/H_{max})$ Column-$II$ (Angle of projection $\theta$)

$A. 1$ $1. 60^o$

$B. 4$ $2. 30^o$

$C. 4\sqrt{3}$ $3. 45^o$

$D. 4/\sqrt{3}$ $4. \tan^{-1}(4) = 76^o$

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