Question

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 not to scale)

• A. A closed loop starts at h, goes down to h/2, then up to the origin, and back to h. Arrows indicate the direction of the loop.
• B. Two straight lines form a triangle. One line goes from the origin to h/2 on the v-axis. The other line goes from the origin to h on the h-axis. Arrows indicate the direction of the lines.
• C. A curve starts at h, goes down to h/2, then up to the origin. An arrow indicates the direction of the curve.
• D. A curve starts at the origin, goes up to h/2, then down to h. An arrow indicates the direction of the curve.

Answer 1

Answer: (A)

Option (a)

Answer 2

A tennis ball is dropped from height $h$ so that during free fall the speed at any height $y$ (with $y=h$ initially and $y=0$ at the floor) is given by

$$v = -\sqrt{2g(h-y)}.$$

At the floor ($y=0$) its speed becomes

$$v = -\sqrt{2gh}.$$

After the bounce (assumed instantaneous) the ball rebounds with speed

$$\sqrt{2g\left(\tfrac{h}{2}\right)} = \sqrt{gh},$$

and on rising its speed at any height $y$ obeys

$$v = \sqrt{gh - 2gy},$$

reaching zero at $y=\tfrac{h}{2}$.

When we plot velocity $v$ (vertical axis) versus height $y$ (horizontal axis), note that:

• At $y=h$ the ball is at rest (point $(h,0)$).
• On its downward journey from $y=h$ to $y=0$, the velocity is uniquely determined by $v = -\sqrt{2g(h-y)}$.
• At the floor $(y=0)$ there are two distinct values: $v = -\sqrt{2gh}$ (just before impact) and $v = \sqrt{gh}$ (just after rebound).
• Finally, during the upward journey from $y=0$ to $y=\tfrac{h}{2}$, the velocity is $v = \sqrt{gh-2gy}$.

Thus, the complete motion in the $v$-$y$ plane traces a closed loop (taking into account the sudden change at the floor) with two distinct branches corresponding to the downward and upward motions.