Force exerted by water on diver (2nd law of Motion)

[bydesign]

I've been in my physics class for almost two months now, and I'm starting to run into trouble. I think I'll be around here for a while, assuming you guys can help me out ;)

The Question : A high diver of mass 68.4 kg jumps off a board 9.5 m above the water.
The acceleration of gravity is 9.8 m/s2 .
If his downward motion is stopped 5.92 s after he enters the water, what average upward
force did the water exert on him?
Answer in units of N.

Homework Equations

:
v = v0 + at (To determine acceleration
EF (Average Force) = ma

The Attempt at a Solution

:
Well, I first have to find the acceleration of the water on the swimmer :
v = v0 + at
0 = (9.5m*9.8m/s^2) + a (5.92)
-93.1 = a (5.92s)
-15.726 m/s^2 = a

Now I plug that into F = ma
F = ma = (68.4 kg) (-15.726 m/s^2) = -1075.682 N

This looks about right, but when I put it into Quest, it comes out wrong. I've checked my math over twice. What am I doing wrong?

Thanks for the help :)

 

[rock.freak667]

you put t=5.92, but that is for when he now enters the water.

so in v=u+at

we want v=0, but the diver doesn't enter the water with 0 velocity. So you need to find this velocity 'u'. (try using conservation of energy to find this velocity)

 

[bydesign]

Thank you very much - that helped me a bunch ;)