Question

A tennis ball is dropped on to the floor from a height of 4m. It rebounds to a height of 3m. If the ball was in contact with the floor for 0.01 second, what was its average acceleration during contact ?

Answer 1

$$\begin{gathered} \mathrm{Initial}\mathrm{velocity}: \\ \mathrm{u}=\sqrt{2{\mathrm{gh}}_1} \\ =\sqrt{2\times 9.8\times 4} \\ =8.85\mathrm{m}/\mathrm{s} \\ \mathrm{Final}\mathrm{velocity}: \\ \mathrm{v}=-\sqrt{2{\mathrm{gh}}_2} \\ =\sqrt{2\times 9.8\times 3} \\ =-7.67\mathrm{m}/\mathrm{s} \\ \mathrm{Average}\mathrm{acceleration}: \\ \mathrm{a}=\frac{△\mathrm{v}}{\mathrm{t}} \\ =\frac{8.85-\left(-7.67\right)}{0.01} \\ =1652\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$