A projectile crosses two walls of equal heigh | vedclass

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(C) 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 peak of t

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) 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 peak of the trajectory is $t_{peak} = \frac{t_1 + t_2}{2} = \frac{2 + 6}{2} = 4 \ s$. The vertical displacement $H$ at time $t = 2 \ s$ is given by $H = (v_0 \sin 	heta) t - \frac{1}{2} g t^2$. At the peak,the vertical velocity is zero,so $v_y = v_0 \sin 	heta - g t_{peak} = 0$,which gives $v_0 \sin 	heta = g t_{peak} = 10 	imes 4 = 40 \ m/s$ (taking $g = 10 \ m/s^2$). Substituting this into the equation for $H$: $H = (40)(2) - \frac{1}{2} (10)(2)^2$ $H = 80 - 20 = 60 \ m$.

$A$ projectile crosses two walls of equal height $H$ symmetrically as shown. The height of each wall is ........ $m$.

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