A body is dropped from a height of 39.2 m. After it crosses half distance, the acceleration due to gravity ceases to act. The body will hit the ground with K.E. (if m = 2kg) :-

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Solution

$The velocity acquired by the body of mass 2 Kg in falling by a distance \frac{39.2}{2} m is given by, v^{2} = 2 \times 10 \times \frac{39.2}{2} = 392 The kinetic energy attained at the time of hitting the ground is, \frac{1}{2} {mv}^{2} = \frac{1}{2} \times 2 \times 392 = 392 J . This is because when no force acts on the body, they will move with constant velocity acquired at the time of force ceases to act .$

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