Impulse-Momentum Theorem and diving board

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SUMMARY

The discussion centers on calculating the force exerted by water on an 82 kg diver who drops from a height of 3 meters and comes to rest in 0.55 seconds after hitting the water. The initial approach using the formula F = (mVf - mVi) / t was incorrect due to a miscalculation of the diver's velocity. The correct method involves using the equation x = (1/2)at² to determine the diver's velocity just before impact, leading to a force calculation of 820 N based on the correct velocity.

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Homework Statement


An 82 kg man drops from rest on a diving board 3m above the surface of the water and comes to rest 0.55 s after reaching the water. What force does the water exert on him?


Homework Equations


F = mVf - mVi / t


The Attempt at a Solution


I have no idea where to begin since I don't know the diver's velocity. Could I possibly use V = d/t?

EDIT: Ok, so I used V = d/t and got 5.5 m/s for the velocity. Then I used F = mVf - mVi / t, and got 820 N. Is this correct?
 
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I'm not saying that I'm in any way a physics expert or anything but;

What you calculated as the velocity was 3m/0.55s --> which is not at all the velocity of the diver just before he hits the water. I made a quick attempt at this problem, and I would instead use the relationship: x=at2/2, since the acceleration is constant. From this you can find the time it takes for the diver to hit the water--also knowing that x=vt, and therefore v=x/t, you can find the velocity of the diver just before he hits the water, which is also the initial velocity in (mVf-mVi)/t.
 

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