A tennis ball is dropped on to the floor from a height of 9 . 8   m . It rebounds to a height 5 . 0…

The speed of ball just before collision with ground is $v_i^2=0+2g{h}_i$

 $\Rightarrow v_{\mathrm{i}}=\sqrt{2g{h}_{\mathrm{i}}}$

$=\sqrt{2\times 10\times 9.8}$

$=14 \mathrm{m} {\mathrm{s}}^{-1} \left(\mathrm{Downward}\right)$

The speed of ball just after collision is $0=v_f^2-2g{h}_f$

$\Rightarrow v_{\mathrm{f}}=\sqrt{2g{h}_{\mathrm{f}}}$

$=\sqrt{2\times 10\times 5}$

$=10\mathbf{ }\mathrm{m} {\mathrm{s}}^{-1 }\left(\mathrm{Upward}\right)$

Average acceleration of ball is $\left|{\overrightarrow{a}}_{\mathrm{avg}}\right|=\left|\frac{\Delta \overrightarrow{v}}{\Delta t}\right|=\frac{v_f-\left(-v_i\right)}{t}=\frac{10+14}{0.2}=\frac{24}{0.2}=120 \mathrm{m} {\mathrm{s}}^{-2}$.