Force/Energy Problem: Jump from Window

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SUMMARY

A 75 kg man jumping from a height of 1.0 m reaches a speed of 4.43 m/s just before impact with the pavement, calculated using the conservation of mechanical energy principles. The potential energy at the height converts entirely into kinetic energy just before the fall. To determine the average force exerted on him by the ground when he lands, the distance of 0.50 cm (0.005 m) for his footpads is used to calculate the deceleration and subsequently apply Newton's second law.

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  • Study the conservation of mechanical energy in physics
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Question

A 75kg man jumpst from a window 1.0m above the side walk.

a) What is his speed just before his feet strike the pavement?

b) If the man jumps with his knees and ankles locked, the only cushion for his fall is approximately 0.50 cm in the pads of his feet. Calculate the average force exerted on him by the ground in this situation.

Work

a) MEi = MEf

PEi + KEi = PEf + KEf ===> mgh + (.5)mv^2 = mgh + (.5)mv^2

After plugging in values:

736 + 0 = 0 + 37.5v^2 ===> V^2 = 19.6 ===> V = 4.43 m/s

b) I'm not sure how to approach this problem. I assume I'm basically trying to find the normal force of the man...but I don't know what the height really has to do with the problem unless I'm supposed to use energies to relate to the force in some way.
 
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You are given the initial velocity from the previous question, and the distance of acceleration by the thickness of his foodpads. I suppose they want you to calculate the constant acceleration needed for this, and then relate that to the force using Newtons second law.

- Kerbox
 

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