Human Falling Determine Force on Achilles

AI Thread Summary
A 45 kg woman jumping from a height of 2.5 meters experiences significant forces upon landing, particularly on her Achilles tendon. After calculating her final velocity and average deceleration, the average force exerted on her feet to stop her fall is determined to be approximately 1378 Newtons. This force is further adjusted based on the distances from her foot to the talus and from the talus to the Achilles tendon, resulting in a calculated force on the tendon. The discussion emphasizes the importance of understanding the biomechanics involved in landing to minimize injury risk. Accurate calculations are crucial for assessing the forces experienced by the body during such impacts.
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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 suff er 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?
 
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