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

Use energy conservation to relate speed and height. During the fall: \(\frac12 mv^2 = mg(h - y)\). During the rise after the bounce: \(\frac12 mv^2 = mg(\frac h2 - y)\). In both cases \(v^2\) is a linear function of \(y\) with slope \(-2g\), so on a \(v\)-\(y\) plot each leg is half of a parabola lying along the y-axis.
Landing speed \(v_1 = \sqrt{2gh}\) (downward, negative) and take-off speed \(v_2 = \sqrt{gh} = v_1/\sqrt2\) (upward, positive), so the upper arc is smaller and reaches only \(y = h/2\), while the lower arc extends out to \(y = h\) with a vertical tangent at \(v = 0\). The two arcs meet the \(v\)-axis at \(y=0\) on opposite sides, the negative one being longer.
This is the figure of option 3.
Answer: (3) a large lower arc ending at h and a smaller upper arc ending at h/2