A ball of 0.1 kg mass is dropped on a hard floor from a height of...

As per conservation of energy, $\frac{1}{2} m u^2=m g h$$\begin{aligned} \Rightarrow \quad u & =\sqrt{2 g h} \\ & =\sqrt{2 \times 10 \times 0.45} \\ & =\sqrt{9}=3 \mathrm{ m} / \mathrm{s} \end{aligned}$After rebound some of its kinetic energy gets lost and remaining gets converted into Potential energy Hence, $\quad \frac{1}{2} m v^2=m g h^{\prime}\begin{aligned} \Rightarrow \quad v & =\sqrt{2 g h^{\prime}} \\ & =\sqrt{2 \times 10 \times 0.20} \\ & =\sqrt{4}=2 \mathrm{ m} / \mathrm{s} \end{aligned}$$$We know that,$\mathrm{F}=\frac{m v-m u}{t}=\frac{m(v-u)}{t}$Taking sign consideration$\mathrm{F}=\frac{0.1(-2-3)}{0.1}=-5 \mathrm{ N} .$ (-ve sign indicated retarding force)