Force/Energy Problem: Jump from Window

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A 75kg man jumping from a 1.0m height reaches a speed of 4.43 m/s just before hitting the ground, calculated using energy conservation principles. The potential energy at the height converts to kinetic energy as he falls. To find the average force exerted on him by the ground when landing with locked knees and a 0.50 cm cushion, the normal force must be calculated using the initial velocity and the deceleration during the impact. This involves determining the constant acceleration required for the short stopping distance and applying Newton's second law. The discussion emphasizes the relationship between energy conservation and force calculations in fall scenarios.
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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
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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