A tennis ball is dropped on to the floor from a height of 9 . 8m ....

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

$⇒ {v}_{i} = \sqrt{2g{h}_{i}}$

$= \sqrt{2 \times 10 \times 9 . 8}$

$= 14m{s}^{- 1}\left(Downward\right)</p><p>The speed$ of ball just after collision is0 = {v}_{f}^{2} - 2g{h}_{f}$</p><p>$⇒ {v}_{f} = \sqrt{2g{h}_{f}}$</p><p>$= \sqrt{2 \times 10 \times 5}$</p><p>$= 10m{s}^{- 1}\left(Upward\right)$</p><p>Average acceleration of ball is \left|\overset{\to}{a}avg\right| = \left|\dfrac{\Delta\overset{\to}{v}}{\Deltat}\right| = \dfrac{{v}_{f} - \left(- {v}_{i}\right)}{t} = \dfrac{10 + 14}{0 . 2} = \dfrac{24}{0 . 2} = 120m{s}^{- 2}$.