A projectile crosses two walls of equal heigh | vedclass

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(C) Let the projectile be thrown with velocity $v_0$ at an angle $	heta$ with the horizontal.The projectile crosses the two walls at times $t_1 = 2 \

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$50$

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(C) Let the projectile be thrown with velocity $v_0$ at an angle $	heta$ with the horizontal.The projectile crosses the two walls at times $t_1 = 2 \ s$ and $t_2 = 6 \ s$.Since the motion is symmetric,the time taken to reach the maximum height is $t_{max} = \frac{t_1 + t_2}{2} = \frac{2 + 6}{2} = 4 \ s$.At maximum height,the vertical component of velocity is zero,so $v_y = v_0 \sin 	heta - gt = 0$.Using $g = 10 \ ms^{-2}$,we get $v_0 \sin 	heta = 10 	imes 4 = 40 \ ms^{-1}$.The horizontal distance between the two walls is $d = 120 \ m$,and the time taken to travel this distance is $\Delta t = t_2 - t_1 = 6 - 2 = 4 \ s$.The horizontal component of velocity is constant: $v_x = v_0 \cos 	heta = \frac{d}{\Delta t} = \frac{120}{4} = 30 \ ms^{-1}$.The magnitude of the projection velocity is $v_0 = \sqrt{(v_x)^2 + (v_y \sin 	heta)^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \ ms^{-1}$.

$A$ projectile crosses two walls of equal height $H$ symmetrically as shown. The velocity of projection is........ $ms^{-1}$.

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