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Step-by-Step Solution
Step 1: Understand the Motion of the Ball
A tennis ball is dropped from an initial height $h$. It falls freely under the influence
of gravity, hits the wooden floor, and rebounds to a height $\tfrac{h}{2}$. Throughout
this motion, the acceleration due to gravity remains constant ($g$ downwards).
Step 2: Relate Velocity to Height while Falling
When dropping the ball from rest at height $h$, just before hitting the floor
(at height $0$), its velocity $v$ can be found from the energy conservation or
kinematics:
$ v = \sqrt{2gh} \quad \text{(downward)}.$
We conventionally take downward velocity as negative and upward velocity as positive
for this type of vertical motion, so we may write $v = -\sqrt{2gh}$ for the
downward motion.
Step 3: Relate Velocity to Height while Rebounding
After the collision with the floor, the ball rebounds and reaches only $\tfrac{h}{2}$.
At its maximum rebound height, the ball’s velocity becomes zero again, but at
height $\tfrac{h}{2}$. As it moves upward, its velocity is positive.
Step 4: Shape of the Velocity–Height Curve
Because the acceleration (i.e., $-g$) is constant, the relationship between velocity
and height becomes a parabolic curve when plotted. Specifically:
The downward motion corresponds to a branch of the parabola in the fourth quadrant
(velocity negative, height decreasing from $h$ to $0$).
The upward motion corresponds to a branch of the parabola in the first quadrant
(velocity positive, height increasing from $0$ to $\tfrac{h}{2}$).
Thus, the correct graph would show a smooth parabolic path passing from the fourth
quadrant (negative velocity) and then into the first quadrant (positive velocity).
This matches the provided correct answer image.
Step 5: Final Explanation
The velocity–height ($v$–$h$) curve is parabolic because the motion is uniformly
accelerated (constant acceleration $-g$). When the ball is falling, $v$ is negative,
and when the ball is rising, $v$ is positive. The heights where $v=0$ occur at $h$
(on release) and at $\tfrac{h}{2}$ (when the ball reaches its rebound peak).
Hence, the curve dips into the fourth quadrant during descent and rises in the
first quadrant during ascent.
