Calculate Distance Buoy Sinks w/ Man on Top

[alexfloo]

Homework Statement

A 810-kg cylindrical can buoy floats vertically in salt water. The diameter of the buoy is 0.900 m. Calculate the additional distance the buoy will sink when a 69.0-kg man stands on top.

Homework Equations

B = ρ{water}Vg

The Attempt at a Solution

I've tried this two ways:
1.) I set up B = Wgt of the buoy, with V = πr2L, and solved for L. Then I set up the same problem with the weight of the buoy plus that of the man and solved again, then subtracted.

2.) The man's weight is equal to the weight of some volume of water. If that water were arranged into a cylindrical formation with radius r = .45m, it would have some length L. This should be equal to the distance the buoy sinks.

A = πr2 = .6362
V = AL = .6362L
ρ(water)V = 636.2L = mass of water

636.2Lg = 69g
L = 69/636.2 = .1085

This is exactly the same result I got for the first same method (as I would expect) but it's apparently wrong. Even more confusing, the web-homework system advised me:

"Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures."

While the numbers listed here are abbreviated, I did the work on my calculator and never actually rounded anything. Any ideas as to what I'm missing?

 

[alexfloo]

Haha okay, I've been staring at this for at least an hour but I just realized the buoy floats in SALT water. The density I used was just slightly to low.

 

[gneill]

Salt water is slightly more dense than pure water. This may be contributing to the problem.

EDIT: Okay, you got there just ahead of my suggestion! Well done.