A projectile crosses two walls of equal heigh | vedclass

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(B) The vertical displacement of the projectile is given by $y = u \sin 	heta t - \frac{1}{2} g t^2$. At $t_1 = 2 \, s$ and $t_2 = 6 \, s$,the projec

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$tan^{-1}(4/3)$

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(B) The vertical displacement of the projectile is given by $y = u \sin 	heta t - \frac{1}{2} g t^2$. At $t_1 = 2 \, s$ and $t_2 = 6 \, s$,the projectile is at the same height $H$. Thus,$H = u \sin 	heta (2) - \frac{1}{2} g (2)^2 = 2 u \sin 	heta - 2g$ and $H = u \sin 	heta (6) - \frac{1}{2} g (6)^2 = 6 u \sin 	heta - 18g$. Equating the two expressions for $H$: $2 u \sin 	heta - 2g = 6 u \sin 	heta - 18g$ $4 u \sin 	heta = 16g$ $u \sin 	heta = 4g = 40 \, m/s$ (assuming $g = 10 \, m/s^2$). The horizontal distance between the two walls is $d = 120 \, m$. The time interval between crossing the two walls is $\Delta t = 6 - 2 = 4 \, s$. Since horizontal velocity $u \cos 	heta$ is constant: $u \cos 	heta = \frac{d}{\Delta t} = \frac{120}{4} = 30 \, m/s$. Now,$	an 	heta = \frac{u \sin 	heta}{u \cos 	heta} = \frac{40}{30} = \frac{4}{3}$. Therefore,$	heta = tan^{-1}(4/3)$.

$A$ projectile crosses two walls of equal height $H$ symmetrically as shown. The angle of projection of the projectile is:

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