Question

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.

Answer 1

i calculate the speed of the diver when she hits the water v = sqrt (2.g.h) = 7,64m/s

Answer 2

I got it:
then we have the equation for the distance travelled from the water surface to the point in the water where the diver comes to a halt:
\[s=v _{0}.t - a.t²/2\]
s is the distance travelled v0 the initial speed at the surface (calculated above = 7,64m/s)
and a.t²/2 the decreasing factor to account for a linear decelerating motion.
this last factor can be rewritten as follows:
\[-a.t²/2 = - \left( F \over m \right).t²/2 = - \left( k.v \over m \right).t²/2\]
because F=m.a and F = k.v (the force F is direct proportional to the speed of travel in the water.
One has this equation and fills in the boundery conditions:
at the point where the body comes to a halt in the water at depth 1.14m, the speed is 0, and thus we get for this last factor 0.
What stays is :
\[s=v _{0}.t \]
or
1,14m = 7,64m/s * t
we get t = 6.70175seconds
Then one can calculate the relation of impuls to force and the time (we can do this since we calculate the average force F over the time t
\[m.v = F.dt\]
thus Faverage = m.v/dt = 64kg.7,64m/s / 6,70175 m/s = 72.96 N