A Tennis ball is released from a height h and after freely falling on a wooden floor it rebounds and reaches height h/2. The velocity versus height of the ball during its motion may be represented graphically by :
(graph are drawn schematically and on not to scale)

Solution

Physics sir · 3 steps

Use v² = 2g × (drop from the top): each leg of the motion is a sideways parabola, not a line.

1. Sign convention. Take upward as positive and let y be the height above the floor. On the way down v is negative. After the bounce v is positive.

   down: v = -√2g(h-y), up: v = +√2g(h/2-y)

2. Shape of each leg. Both legs follow v² ∝ (top height - y), which is a parabola lying on its side. At each turning point (y = h and y = h/2) the curve meets the height axis vertically. The speed is largest at the floor: √2gh just before impact and √gh just after.

   Option	What the graph shows	Verdict
   (a)	Straight lines, so v changes in proportion to height	wrong, v goes as a square root
   (b)	Flat at the turning points, speed shooting up without limit near the floor	wrong, speed at the floor is finite
   (c)	Negative parabolic arc from h to the floor, jump, then positive arc up to h/2	matches
   (d)	Speed falls off as the ball nears the floor and grows from zero just after the bounce	wrong, speed is largest at the floor

3. So. Only (c) has the right signs, the parabolic shape and the largest speed at the floor. The answer is (c).

The idea to keep. For velocity against position under constant acceleration, use v² = u² + 2as. You get a sideways parabola that cuts the position axis vertically wherever the body turns back.

Answer from the NTA final answer key. Solution by PrepWiser.