A64.0-kg person jumps from rest off a 2.98-m-high tower straight down into the water. neglect air resistance. she comes to rest 1.14 m under the surface of the water. determine the magnitude of the average force that the water exerts on the diver. this force is nonconservative.

The solution to the problem is as follows:

person's speed when water is hit = V = √2gh = 7.65 m/s 

person's momentum when water is hit = 7.65m = 489.6 kgm/s 

avg speed under water = 7.65/2 = 3.825 m/s 

time to stop under water = 1.14/3.825 = 0.30 s 

Favg = change in momentum/time to stop = 489.6/0.30 ≈ 1632 N ANS


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