Jumping Forces?

Question

An exceptional standing jump would raise a person 0.80m off the ground. To do this, what force must a 66-kg person exert against the ground? Assume the person crouches at a distance of 0.20m prior to jumping, and thus the upward force has this distance to act over before he leaves the ground.

Answer 1

We know thta height for a free fall is
h=.5gt^2
h-height
g- gravitational acceleration
t- time

We know that (Average power)
Power (P)=change in energy (E) over time (t)
and
Power (P)= force (F) times velocity (v)
We can write
P(avg)=(E2-E1)/(t2-t1)
E2=mgh2
E1=mgh1
h2-h1=.5gt2^2
t1=0
So
(E2-E1)/(t2-t1) = Fv
Oh yeas v(avg)=g(t2-t1) or v=gt2
we have
(E2-E1)/(t2)= Fgt2
then F=(E2-E1)/[g(t2)^2]
and substituting E=mgh we have

F=mg(h2-h1)/[g(t2)^2]
since h2-h1=.5gt2^2 we have

F=mg(.5gt2^2)/[g(t2)^2]
F=.5mg
F=.5 x 66 x 9.81=323.73N

The total force exerted on the ground however will be the weight of the person pluss the force
Ft=1.5 x x 66 x 9.81 = 971 N

Let me know if it makes sense.