- #1

smitty13

- 1

- 0

## 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?