A projectile crosses two walls of equal heigh | vedclass

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(B) The projectile crosses the two walls at $t_1 = 2 \ s$ and $t_2 = 6 \ s$. Due to symmetry,the time taken to reach the maximum height is $T_{max} =

Vedclass · Accepted Answer

$80$

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(B) The projectile crosses the two walls at $t_1 = 2 \ s$ and $t_2 = 6 \ s$. Due to symmetry,the time taken to reach the maximum height is $T_{max} = \frac{t_1 + t_2}{2} = \frac{2 + 6}{2} = 4 \ s$. Using the equation of motion for vertical displacement,$y = u_y t - \frac{1}{2} g t^2$. At $t = 4 \ s$,the vertical velocity $v_y = 0$. Using $v_y = u_y - gt$,we get $0 = u_y - 10(4)$,so $u_y = 40 \ m/s$. The maximum height $H_{max}$ is given by $H_{max} = \frac{u_y^2}{2g} = \frac{40^2}{2 	imes 10} = \frac{1600}{20} = 80 \ m$.

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

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