An average person can reach a maximum height of about 60 cm when jumping straight up from a crouched position. during the jump itself, the person's body from the knees up typically rises a distance of around 50 cm. to keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump.

1) You can find the velocity with which the jumper leaves the ground by using the formula for the final velocity of an upward vertical motion.

Vf^2 = Vo^2 - 2gd

The data known are: Vf = 0, g = 9.81 m/s^2,   and d = 60 cm = 0.6 m

=> Vo^2 = 2gd = 2*9.81m/s^2 * 0.6 m = .11.772 m^2/s^2 =>

Vo = 3.4 m/s

2) You can obtain the acceleration of the jumper during the jump by using the equivalent formula, Vf^2 = Vo^2 + 2ad

Where, Vf^2 is the velocity with which he leaves the ground (3.4 m/s), Vo is zero because he jumps from the rest, d is the distance that his body rises from knees,which is 50 cm = 0.5 m

Then,  a = Vf^2 / 2d = [3.4m/s]^2 / (2*0.5m) =   11.6 m/s^2

And now, you can find the force exerted by the jumper as:

F = m*a, and if you know his weight, W = m*g => m = W/g

Replace m in F = m*a => F = W*a/g = W*11.6m/s^2 / 9.81 m/s^2 = 0.10 W

F = 0.10 W 

 

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j