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(C) Let the hero and the villain be at the same horizontal level at $t = 0$. At this instant,the villain fires the bullet horizontally.Both the hero a

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The bullet will pass below the hero.

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(C) Let the hero and the villain be at the same horizontal level at $t = 0$. At this instant,the villain fires the bullet horizontally.Both the hero and the villain have a downward acceleration of $a_p = 2\ m/s^2$.Since the bullet is fired horizontally,its initial vertical velocity is $0$. However,once the bullet is in the air,it is only subject to gravity (ignoring air resistance),so its vertical acceleration is $a_b = g = 10\ m/s^2$ (downwards).Let $t$ be the time taken for the bullet to reach the hero's horizontal position.The vertical displacement of the hero in time $t$ is $y_h = u_y t + \frac{1}{2} a_p t^2 = u_y t + \frac{1}{2} (2) t^2 = u_y t + t^2$.The vertical displacement of the bullet in time $t$ is $y_b = u_y t + \frac{1}{2} a_b t^2 = u_y t + \frac{1}{2} (10) t^2 = u_y t + 5t^2$.Since $5t^2 > t^2$,the bullet will have a greater downward displacement than the hero in the same amount of time.Therefore,the bullet will pass below the hero.

Column $I$	Column $II$
$(A)$ Horizontal component of velocity	$(p)$ $5$ $SI$ unit
$(B)$ Vertical component of velocity	$(q)$ $10$ $SI$ unit
$(C)$ Horizontal displacement	$(r)$ $15$ $SI$ unit
$(D)$ Vertical displacement	$(s)$ $20$ $SI$ unit

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