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A 64.0-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. ...?

Respuesta :

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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