DFQ: "A vertical conical tank..."

Hey guys.

I've spent probably 6 hours today trying to solve a basic differential equation, without success, and I'm about to go crazy.

The problem:

"A vertical conical tank, point down, with a radius of 0.5 meters, and a height of 3 meters, is filled with water. How long does it take to drain the tank through a small hole in the bottom of the tank, when the water level has sunk by 1.5 meters after 45 minutes?"

The answer is supposed to be ~55 minutes.

I'm not gonna do the entire thing here, but here is general way of how I tried solving it:

1) Volume of a cone is $\displaystyle V = \frac{\pi * r^2 * h}{3}$ => $\displaystyle V(t) = \frac{\pi * r^2}{3} * h(t)$

2) General solution to a "draining tank through a small hole" is $\displaystyle \frac{dV}{dt} = -k * \sqrt{h}$

3) Rearranging and finding the derivative, getting $\displaystyle \frac{\pi * r^2}{3} * \frac{dh}{\sqrt{h}}= -k dt$

4) Integrating both sides = $\displaystyle \frac{2\pi * r^2}{3} * \sqrt{h} = -kt+C$

5) Finding an expression for h(t): $\displaystyle h(t) = ((-kt+C) * \frac{3}{2\pi*r^2})^2$

6) Using t = 0 to find C, and then the information about the water level after 45 minutes, to find k.

7) Finding the time when h(t) = 0 (tank is empty). Standard second degree equation.

8) Getting the wrong answer. All but one time, I've gotten t = 2.56 hours, which is completely wrong.

Anyone able to see if there's anything wrong with that? Or am I just making some sort of fluke error while calculating? Anyone able to get it right (~55 minutes)?

Thanks!