A Tennis ball is released from a height h and after freely falling on a wooden floor it rebounds and reaches height
Question:
A Tennis ball is released from a height $\mathrm{h}$ 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)
Correct Option: , 3
Solution:
(3) For uniformly accelerated/ deaccelerated motion :
$v^{2}=u^{2} \pm 2 g h$
As equation is quadratic, so, $v-h$ graph will be a parabola
Initially velocity is downwards (-ve) and then after collision it reverses its direction with lesser magnitude, i.e. velocity is upwards (+ve).
Note that time $t=0$ corresponds to the point on the graph where $h=d$.
Next time collision takes place at 3 .