An 70 kg man drops from rest on a diving

board 3.2 m above the surface of the water
and comes to rest 0.57 s after reaching the
water.

The acceleration due to gravity is
9.81 m/s2.

What force does the water exert on the
man? Answer in units of N.

4 answers

avgforce*time=change in momentum
Calculate the velocity at impact (free fall, 3.2m), multiply by mass.
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