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(C) Let the initial velocity of the bullet be $u$ and the angle of projection be $	heta$. The target is at a distance $d$ from the gun. The time take

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Hit the target

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(C) Let the initial velocity of the bullet be $u$ and the angle of projection be $	heta$. The target is at a distance $d$ from the gun. The time taken for the bullet to reach the horizontal distance $d$ is $t = d / (u \cos 	heta)$.In this time $t$,the vertical displacement of the bullet due to gravity is $y_1 = (1/2) g t^2$.The vertical position of the bullet relative to the line of sight (which is at angle $	heta$) is $y_2 = (u \sin 	heta) t - (1/2) g t^2$. Since the target is in the line of the barrel,the initial height of the target is $h = d 	an 	heta = (u \sin 	heta) t$. The actual position of the target at time $t$ is $h' = h - (1/2) g t^2 = (u \sin 	heta) t - (1/2) g t^2$.Since the vertical position of the bullet $y_2$ is equal to the vertical position of the target $h'$,the bullet will hit the target.

$A$ gun is aimed at a target in the line of its barrel. The target is released and allowed to fall under gravity at the same instant the gun is fired. The bullet will

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