Question

1. The ball of mass 2kg is thrown upwards at a height of 4m.When comming down its initial velocity is 3m/s. What will be its kinetic energy just before hitting the ground ?

Answer 1

$$\begin{gathered} \mathrm{Given}, \\ \mathrm{m}=2\mathrm{kg} \\ \mathrm{u}=3\mathrm{m}/\mathrm{s} \\ \mathrm{h}=4\mathrm{m} \\ \mathrm{Using}\mathrm{relation}, \\ {\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{gh} \\ {\mathrm{v}}^2-9=2\times 9.8\times 4 \\ \mathrm{or}\mathrm{v}=8.3\mathrm{m}/\mathrm{s} \\ \mathrm{Now},\mathrm{kinetic}\mathrm{energy}=\frac{1}{2}{\mathrm{mv}}^2 \\ =\frac{1}{2}\times 2\times 8.3^2=68.89\mathrm{J} \end{gathered}$$