How High Must a Diver Jump to Experience a Force Five Times Their Weight?

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avb203796
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A 50.0 kg person dives from a cliff above the water. From what height must he dive if he experiences an average force equal to 5.0 times his body weight (W=mg), as he comes to a stop in the water? You may assume a stopping time of 0.400 s.

So first I solved for the average force:

average F= 5.0(mg) = 5.0(9.80m/s^2)(50.0kg) = 2450N

Next I used the equation m*delta v = average F*delta t to solve for delta v:

delta v = [(2450)(0.4)]/50.kg = 19.6 m/s

I was wondering if my above calculations are right and then I am not sure how to now find the height using the delta v!
 
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ok .. now that you have found out delta v, you know the man's velocity as he reaches the water ... use the equations of motion to find the ht.

(you can use conservation of energy too)
 
Shouldn't my delta v be negative though because delta v = final velocity - intial velocity and my final velocity would be 0 because the diver is coming to a stop?
 
avb203796 said:
Shouldn't my delta v be negative though because delta v = final velocity - intial velocity and my final velocity would be 0 because the diver is coming to a stop?

Check the signs on your vectors. That will explain it.

-Dan