Question

A Tennis ball is released from a height \(h\) and after freely falling on a wooden floor it rebounds and reaches height \(\frac{h}{2}\). The velocity versus height of the ball during its motion may be represented graphically by :
(graph are drawn schematically and on not to scale)

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

Use \(v^2 = 2g(H - y)\) on each leg: the fall gives a curve from (h, 0) to a large negative v at height 0, and the rebound gives a smaller positive v at height 0 falling to zero at h/2.

Answer 1

The Correct Option is C

Take upward as positive and plot \(v\) against height \(y\).
Fall from \(h\): \(v^2 = 2g(h - y)\), so \(v = -\sqrt{2g(h-y)}\). The curve starts at \(y = h\) with \(v = 0\) and reaches \(v = -\sqrt{2gh}\) at \(y = 0\): a parabolic arc with a vertical tangent at \(y = h\).
Rebound: the ball leaves the floor with \(v = +\sqrt{2g(h/2)} = +\sqrt{gh}\), which is smaller than the impact speed, and rises with \(v = +\sqrt{2g(h/2 - y)}\), reaching \(v = 0\) at \(y = h/2\).
So the graph has a larger lower arc (negative v) running between \(y = h\) and the v-axis, and a smaller upper arc (positive v) running between the v-axis and \(y = h/2\); the arrow on the lower arc points from h towards the v-axis and the arrow on the upper arc points from the v-axis towards h/2. This is the third figure.
Option 1 shows only the lower branch with a wrong shape, option 2 has straight lines (v would be linear in y only if the acceleration were not constant), and option 4 is a closed loop that does not reach the v-axis at the larger speed.
Answer: (3) a large lower arc ending at h and a smaller upper arc ending at h/2