A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after…

When ball of mass $\text{'}\mathrm{m}\text{'}$ dropped from height and strikes to ground.It's kinetic energy is converting into elastic potential energy due to compression.

So after time $\text{'}\mathrm{t}\text{'}$ its final velocity,

$\mathrm{v}=\mathrm{u}+\mathrm{at}$

$\mathrm{u}$ is initial velocity of ball, $\mathrm{a}$ is acceleration of ball and $\mathrm{t}$ is time.

$\Rightarrow \mathrm{v}=0+\left(-\mathrm{g}\right)\mathrm{t}$

$\mathrm{v}=-\mathrm{gt}$

Negative sign shows downward direction.

Now kinetic energy of ball,

$\mathrm{K}.\mathrm{E}.=\frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}\mathrm{m}{\left(-\mathrm{gt}\right)}^2$

$\Rightarrow \mathrm{K}.\mathrm{E}.=\frac{1}{2}{\mathrm{mg}}^2{\mathrm{t}}^2$

$\Rightarrow \mathrm{K}.\mathrm{E}.\propto {\mathrm{t}}^2$

So, it is upward parabola. But after complete compression, $\mathrm{K}.\mathrm{E}.$ becomes zero, then ball will reshape again. Kinetic energy will not be constant.