A body of mass 2 kg is projected vertically upwards with a velocity of $2\mathrm{m}{\sec }^{-1}$. The K.E. of the body just before striking the ground is (in J)

If particle is projected vertically upward with velocity of 2m/s then it returns with the same velocity. 

So its kinetic energy $=\frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}\times 2\times {\left(2\right)}^2=4\mathrm{J}$