An 82 kg man drops from rest on a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching the water. What is the net force on the diver as he is brought to rest?

Easiest to use work and energy principals here. Set the work done by gravity as the man falls from a height of +3m to a depth of -.55m equal to the work done by the water to slow him down.

**Correction: this method gives the force exerted by the water. But the question asks for net force, so we must add the effect of gravity operating on the man, so subtract mg.

Perhaps a better way is to use v^2 = 2gh to find his velocity when he hits the water, and then v^2 = 2ad to find his deceleration in the water, then set F=ma.