Question

A Tennis ball is released from a height $h$ and after freely falling on a wooden floor it rebounce and reaches height$\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$. The velocity versus height of the ball during its motion may be represented graphically by: (graphs are drawn schematically and are not to scale)

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Step 1. Given Data,

Height $h$,

Height after free falling$\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$,

Step 2. The velocity $v$ versus height $h$ of the ball during its motion is represented graphically by,

After freely falling on a wooden floor the ball bounce and do a projectile motion.

Hence the velocity $v$ versus height $h$ the $\left(v,h\right)$ curve will be parabolic.

At, Height $H=h,$ the velocity $v=0$
At, Height $H=0,$ the velocity $v=\sqrt{2gh}$
Also $a=–g$, throughout this motion. Then downward velocity is negative and upward is positive.

When the ball is coming down graph will be in $IV$ quadrant that is $v$ is negative and when going up graph will be in$I$ the quadrant that is $v$ is positive.

$\Rightarrow v^2=u^2+2gh$

Hence, the correct option is A.