A man whose mass is `m` kg jumps vertically into air from the 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 at height `h_(2)` from the ground and finally centre of mass rises to `h_(3)` above the ground when he is at the top of the jump. what is the average upward force exerted by the ground on him?

Options

A. `(mg(h_(3)-h_(1)))/((h_(3)-h_(2)))`
B. `(mg(h_(3)-h_(1)))/(h_(3))`
C. `(mg(h_(3)-h_(1)))/((h_(2)-h_(1)))`
D. none of these

Answer

C

Explanation

From `h_1 to h_2`, let the average force be F. Acceleration a will be upwards.
`F-mg=maimpliesa=(F-mg)/m`
Let v be the velocity of centre of mass when the contact is left:
`v^2=2a(h_2-h_1)`
`impliesv^2=2((F-mg)/m)(h_2-h_1)`

Now he is in free fall. Further the height raised `=h_3-h_2`.
So, `h_3-h_2=v^2/2g`
`impliesh_3-h_2=((F-mg)/(mg))(h_2-h_1)`
`implies F=(mg(h_3-h_1))/((h_2-h_1))`