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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
Explanation
Before hitting the ground, the velocity v is given by $v^{2} = 2gd$ Further, $v'^{2} = 2g \times (d/2) = gd$ $\therefore (v/v') = \sqrt{2}$ or $v = v'\sqrt{2}$ As the direction is reversed and speed is decreased and hence graph (a) represents these conditions correctly.