1. Motion during the fall

A ball is released from rest from height $h$.

Using $v^2 = u^2 + 2as$ for downward motion:

• initial velocity $u=0$
• acceleration magnitude $=g$
• distance fallen from the release point to a point at height $y$ above the floor is $(h-y)$

So, $v^2 = 2g(h-y)$ Hence the speed is $v = \sqrt{2g(h-y)}$ if we plot speed vs height.

If velocity is taken with upward positive sign, then during downward motion $v = -\sqrt{2g(h-y)}$ So as height decreases from $h$ to $0$, velocity goes from $0$ to $-\sqrt{2gh}$ This is one curved branch below the height-axis reference line for $v=0$.

2. Motion after rebound

After hitting the floor, the ball rises to height $\dfrac h2$.

Therefore, its speed just after rebound is found from $0 = u^2 - 2g\left(\frac h2\right)$ so $u^2 = gh$ $u = \sqrt{gh}$

Thus at height $y$ during upward motion, $v^2 = gh - 2gy$ so $v = +\sqrt{g(h-2y)}$ for $0\le y\le \frac h2$

This branch starts at the floor with positive velocity $+\sqrt{gh}$ and decreases to $0$ at height $\dfrac h2$.

3. Key graphical features

So the $v$ vs height graph must have:

1. A negative branch from: $(y=h,\ v=0) \to (y=0,\ v=-\sqrt{2gh})$
2. A sudden jump at the floor ($y=0$) because collision changes velocity instantaneously from $-\sqrt{2gh} \to +\sqrt{gh}$
3. A positive branch after rebound from: $(y=0,\ v=+\sqrt{gh}) \to \left(y=\frac h2,\ v=0\right)$

Also note:

• the downward speed at the floor is larger than the upward speed after rebound: $\sqrt{2gh} > \sqrt{gh}$
• both branches are curved (square-root type), not straight lines.

4. Matching with options

The correct schematic must show:

• lower negative branch ending at floor with larger magnitude,
• upper positive branch starting at floor with smaller magnitude,
• upper branch reaching $v=0$ at height $h/2$,
• lower branch reaching $v=0$ at height $h$.

This corresponds to Option C.

5. Comparison with stored answer

Stored correct answer: C

My derived answer: C

So they agree.