**an 82 kg man drops from rest on a diving board 3.0 m above the surface of the water and ocmes to rest 0.55 s after reaching the water. what force does the water exert on him?**

**Answer:**

If g=9.8 is the acceleration of gravity, we have (1/2)*g*t^2=3, that is t^2=6/g, which means t=0.78

This is the time needed for the man to reach the water. Therefore his speed v at the time is equal to

g*0.78=7.64 . Therefore the deceleration of the man caused by the water, (assuming it is constant of course) is equal to 7.64/0.55=13.9 . This means that the force F that the water exerts on him must be F=82*13.9=1139 N since F=m*a.