A man whose mass is m kg jumps vertically into air from a sitting position in which his centre of mass is at a height `h_(1)` from the ground. When his feet are just about to leave the ground his centre of mass is `h_(2)` from the ground and finally rises to `h_(3)` when he is at top of the jump . What is the average upwards force exerted by the ground on him ?

Answer

`(mg(h_(3) - h_(1)))/((h_(2) - h_(1)))`

Explanation

From work energy theoem
`W_(F) + W_(g) = DeltaKE`
`F_(avg) (h_(2) - h_(1)) - mg(h_(3) - h_(1)) = 0 rArr F_(avg) = (mg(h_(2) - h_(1)))/((h_(2) - h_(1)))`