A ball is dropped vertically from a height above the ground. It hits the ground and bounces up vertically to a height . Neglecting subsequent motion and air resistance, its velocity varies with height above the ground as
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: A
Solution
During the fall from height d, the ball accelerates downward, so v = √(2g(d−h)), meaning v decreases linearly with h² — it's a parabola opening rightward with v=0 at h=d and maximum v at h=0. After bouncing, the ball decelerates going upward, reaching only h=d/2, so v = √(2gh), another parabola but now v=0 at h=0 and maximum at h=0 (ground). The key features are: (1) the graph is parabolic (v proportional to √h or √(d−h)), not linear, (2) the velocity is discontinuous at h=0 because the ball hits with speed √(2gd) but bounces back with a lower speed √(gd) (since it only reaches d/2), and (3) the upward branch ends at h=d/2 while the downward branch starts at h=d. Option 1 correctly shows two separate parabolic curves meeting the h-axis at different points (h=d going down, h=d/2 going up) with a discontinuity in v at h=0, reflecting energy loss on impact.
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