A physics Problem?

Question

A 65 kg swimmer jumps off a 10m tower.
A. Find the swimmer's velocity on hitting the water. (I got this because I understood this question. It came out to 14m/s)
B. The Swimmer comes to a stop 2.0m below the surface. Find the net force exerted by the water.

I tried this, Vi=14m/s, Vf=0, d=12m, I thought about using Vf^2=Vi^2+2ad where 0=(14m/s)^2+ (2(a)(12m) How can I rearrange the problem so that I can solve for a? Or am I using the wrong formula???

Answer 1

Use the same formula that you used to determine the speed on hitting the water, but this time, distance = 2m. Therefore v² = 2ad, so a = v²/2d = 196/4 = 49m/s². Force = mass x acceleration = 65 x 49 = 3185 newtons.

Answer 2

v at the surface v^2 = 2 g h effectively 14 m/s

the speed decreases with v = at + v0 (1) where a is negative
so you have t = (v-v0)/a (2)

if you integer (1) you have x = at^2 /2 +vot

replacing t by (2) and x=2

2 = a* (v-v0)^2 /2a + v0 (v-v0)/a

You find (verify) (v0) ^2 = a

and f = ma