Conservation of mechanical energy

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Question

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

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Solution

If particle is projected vertically upward with velocity of 2m/s then it returns with the same velocity.
So its kinetic energy $= \frac{1}{2} {mv}^{2} = \frac{1}{2} \times 2 \times {(2)}^{2} = 4 J$

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