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(A) For a ball dropped from height $d$,the velocity $v$ at any height $h$ is given by $v^2 = u^2 + 2a(s)$. Here,$u = 0$ and $a = -g$,so $v^2 = -2g(h -

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(A) For a ball dropped from height $d$,the velocity $v$ at any height $h$ is given by $v^2 = u^2 + 2a(s)$. Here,$u = 0$ and $a = -g$,so $v^2 = -2g(h - d) = 2g(d - h)$. This implies $v = \pm \sqrt{2g(d - h)}$.$1$. During the downward motion from $h = d$ to $h = 0$,the velocity is negative (downward) and its magnitude increases as $h$ decreases. The relation $v = -\sqrt{2g(d - h)}$ represents a parabolic curve opening towards the positive $h$-axis.$2$. At $h = 0$,the ball hits the ground. Just after the collision,it bounces up to a height $d/2$. The velocity becomes positive (upward) and its magnitude is determined by $v = \sqrt{2g(d/2 - h)}$.$3$. As the ball moves upward from $h = 0$ to $h = d/2$,the velocity decreases from its maximum value to zero at $h = d/2$. This also follows a parabolic path.Comparing these physical requirements with the given options,the graph that correctly represents the downward motion (negative velocity) and the subsequent upward motion (positive velocity) is option $(A)$.

$A$ ball is dropped vertically from a height $d$ above the ground. It hits the ground and bounces up vertically to a height $d/2$. Neglecting subsequent motion and air resistance,its velocity $v$ varies with the height $h$ above the ground as:

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