A ball dropped from a building of height 12m falls on a slab of 1 m height from the ground and makes a perfect elastic collision. Later the ball falls on a wooden table of height 0.5m, makes inelastic collision and falls on the ground. If the coefficient of restitution between the ball and the table is 0.5, then the
velocity (ms^-1) of the ball while touching the ground is about (g = 10 ms^–2) $\_\_\_\_\_$ × 10^–1.

From the question,

the ball is dropped from point P, and collides at point Q from the question by third equation of motion,

${{\mathrm{v}}_{\mathrm{Q}}}^2={\mathrm{u}}^2+2\mathrm{gh}=0+2\times 10\times (12-1)=220$

${\mathrm{v}}_{\mathrm{Q}}=\sqrt{220}\mathrm{m}/\mathrm{s}$

There is no kinetic energy lost in a point Q perfect elastic collision.
Thus, the ball's height after impact

$$\begin{gathered} \frac{1}{2}\mathrm{m}{{\mathrm{v}}_{\mathrm{Q}}}^2=\mathrm{mgh} \\ \Rightarrow \frac{1}{2}\times {\left(\sqrt{220}\right)}^2=10\mathrm{h} \\ \Rightarrow \mathrm{h}=11\mathrm{m} \end{gathered}$$

Using the energy conservation law, let the ball's velocity before an elastic contact be equal to $v_s$

$\frac{1}{2}\mathrm{m}{{\mathrm{v}}_{\mathrm{s}}}^2=\mathrm{mg}(11+0.5)$

${{\mathrm{v}}_{\mathrm{s}}}^2=2\times 10\times 11.5$

${\mathrm{v}}_{\mathrm{s}}=\sqrt{230}\mathrm{m}/\mathrm{s}$

$\mathrm{e}=0.5=\frac{\mathrm{velocity} \mathrm{after} \mathrm{collision} \left(\mathrm{v}\right)}{\mathrm{velcoity} \mathrm{before} \mathrm{collision} \left({\mathrm{v}}_{\mathrm{s}}\right)}$

$$\begin{gathered} 0.5=\frac{\mathrm{v}}{\sqrt{230}} \\ \Rightarrow \mathrm{v}=0.5\sqrt{230}\mathrm{m}/\mathrm{s} \end{gathered}$$

By energy conservation law,

$\frac{1}{2}{\mathrm{mv}}^2=\mathrm{mgh}$

$$\begin{gathered} \frac{1}{2}{(0.5\sqrt{230})}^2=10\mathrm{h} \\ \Rightarrow \mathrm{h}=\frac{0.5\times 230\times 0.5}{20}=2.875\mathrm{m} \end{gathered}$$

Thus the total height from T,

$\mathrm{h}=\mathrm{h}+0.5=2.875+0.5=3.375$

Energy conservation law,

$\frac{1}{2}\mathrm{m}{{\mathrm{v}}_{\mathrm{r}}}^2=\mathrm{mgh}$

${\mathrm{v}}_{\mathrm{T}}=\sqrt{2\mathrm{gh}}=\sqrt{2\times 10\times 3375}=8.215\mathrm{m}/\mathrm{s}$

Hence the correct answer is 82 x 10^-1.