Good effort! I'm answering this in the morning, so I have an advantage. I think your logic is correct, but in my opinion you've made a couple of incorrect assumptions....
Your assumption that the annulus area at the back of the aircraft is the same as the annulus area at the front will not be true (I don't know what the dimensions are, but they won't be the same. You can tell this just by looking at the engine). Also, you have taken the intake area to be the area of a circle of a diameter equalling the fan diameter. But what about the spinner? The intake area is actually an annulus with an outer diameter equal to the diameter of the fan, and a height equal to the height of the fan blades, significantly less than the fan case diameter. Again, I don't have these figures to hand.
I think the best way of working it out would be to work out the mass flow at the intake (mass flow equals annulus area multiplied by density multiplied by velocity of intake, which is usually designed to be about Mach 0.6), assume that the cabin pressurization bleeds are negligible (obviously some of this mass flow of air that goes in the front won't come out of the back, instead it will end up in the aircraft cabin), and then
use your equation for thrust to work out the average velocities of the hot and cold jets (velocity equals required thrust divided by mass flow). Obviously this is true only at take-off (for other conditions thrust won't be 115klbf, density won't be 1.225kg/m^-3, and we would have the aircraft speed to worry about).
Skipping the first part of the above, according to random website (http://www.turbokart.com/about_ge90.htm), GE90-115 mass flow is approx 3000lb/s, which is 1361kg/s. 115lbf in N is 511.545KN. Therefore, average of the hot and cold jet velocities is 511,545/1361 = 375.8m/s, or 838.8 mph
This speed is only valid while the aircraft is not moving forwards and attempting to create 115Klbf of thrust. If it is moving forwards, add the forward speed of the aircraft to this number. Obviously my value is wrong if the figure of 3000lb/s is incorrect. Also, the cold jet will be moving somewhat slower than the hot jet, which will be significantly faster (Mach 1, I believe, but I don't know the temperature or pressure, so Idon't know what that is in m/s).
[Edited 2006-09-10 11:45:07]