Human Falling Determine Force on Achilles

[smitty13]

Homework Statement

A 45 kg woman jumps from a height of 2.5 meters to a solid
floor. In an attempt to minimize the likelihood of injury, she lands on the ball
of her feet (the pads just behind her toes) with her knees bent. What
average force is needed on her feet to stop her fall if her center-of-gravity
after touching the floor drops 80 cm before she stops? If the horizontal
distance from the ball of her foot to her talus (where her tibia exerts its
force) is 15 cm, and from talus back to the Achilles tendon is 3.6 cm, what
average force does the Achilles tendon suffer during the stopping time

Homework Equations

F = ma
vf^2 = vi^2 + 2ad

The Attempt at a Solution

I've solved for vf = 7 m/s and the average deceleration of the woman (49/(2*.8m)) = 30.625 and even got to an average force answer of 6028.13 Newtons per second using average deceleration and time but just got stuck from there. Can anybody help me out?

 

[Spinnor]

Average force = m*average acceleration = 45*30.625 = 1378 Newtons ?

For the last part see

http://www.google.com/imgres?q=achi...w=1800&h=655&ei=A65mTsCcGYnz0gHz6JCuCg&zoom=1

The 1378 Newton force will be multiplied by a factor of the two distances given

F tendon = 1378 * 15/3.6

?