Calculate the uncertainty in velocity of a cricket ball of mass 150 g if the uncertainty in its position is $1\mathrm{\hspace{0.17em}}\stackrel{\mathrm{o}}{\mathrm{A}}\mathrm{\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{h}=6.6\mathrm{\hspace{0.17em}\hspace{0.17em}}\times \mathrm{\hspace{0.17em}\hspace{0.17em}}{10}^{-34}\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{kg}\mathrm{\hspace{0.17em}\hspace{0.17em}}{\mathrm{m}}^2\mathrm{\hspace{0.17em}}{\mathrm{s}}^{-1}\right)$

$\begin{array}{l}\mathrm{\Delta x}\times \mathrm{m}.\mathrm{\Delta v}\ge \frac{\mathrm{h}}{4\mathrm{\pi }}\\ \mathrm{m}=150\mathrm{g}=0.15\mathrm{kg}\\ \mathrm{\Delta x}=1{\mathrm{A}}^{\mathrm{o}}={10}^{-10}\mathrm{metre}\\ \mathrm{h}=6.6\times {10}^{-34}\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}\\ \mathrm{p}=3.14\\ \mathrm{\Delta V}=\frac{6.6\times {10}^{-34}}{0.15\times {10}^{-10}\times 4\times 3.14}\\ =3.5\times {10}^{-24}{\mathrm{ms}}^{-1}\end{array}$