Physics q--help!?

Question

A 59 kg swimmer jumps off a 9 m tower so....

Find the swimmer's velocity when hitting the water.

The swimmer comes to a stop 2 m below the surface. Find the net force exerted by the water.

Answer 1

a) Potential energy at the top equal to kinetic energy just before the swimmer hits the water
Pe=Ke
mgh=(1/2) m V*2
V= sqrt(2gh)= sqrt(2 x 9.81 x 9.0)
V=13.3 m/s

b )Work done by water W is a product of net force F exerted by water on the swimmer times the distance s.
W=Fs so the net force F=W/s
and since Pe=Ke=W we have
mgh=Fs
F= mgh/s=
F=59 x 9.81 x 9.0/2.0=2600 N

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