Impulse-Momentum Theorem and diving board

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An 82 kg man drops from a diving board 3 meters high and comes to rest 0.55 seconds after hitting the water. To find the force exerted by the water, the diver's velocity just before impact must be calculated, which is not the same as the average velocity during the fall. Using the equation for constant acceleration, the correct approach involves determining the time to hit the water and then using that to find the final velocity. The initial attempt at calculating the force yielded 820 N, but the method used to find the velocity was incorrect. Accurate calculations are essential for determining the correct force exerted by the water on the diver.
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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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