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: (graphs are drawn schematically and on not to scale)

Taking the ground as origin ($$y = 0$$) and upwards as positive:

$$\text{For downward motion from height } h: v^2 = 0 - 2g(y - h) \implies v = -\sqrt{2g(h - y)}$$

This is a parabolic curve in the fourth quadrant ($$v < 0$$) with its vertex at $$(h, 0)$$ that curves towards the $$v$$-axis as height decreases to zero.

$$\text{For upward motion after rebound to height } \frac{h}{2}: 0 = v_0^2 - 2g\left(\frac{h}{2} - 0\right) \implies v^2 = 2g\left(\frac{h}{2} - y\right) \implies v = +\sqrt{2g\left(\frac{h}{2} - y\right)}$$

This is a parabolic curve in the first quadrant ($$v > 0$$) with its vertex at $$\left(\frac{h}{2}, 0\right)$$ where velocity decreases as height increases.

Answer: Option (C): The graph consists of two parabolic segments opening towards the negative $$h$$-axis with vertices at $$h$$ and $$\frac{h}{2}$$.