How far can an aircraft carrier catapult fling a four-ton pickup truck?

pickup-truck-aircraft-carrier-catapult

How far can a pickup truck be flung off of an aircraft carrier?

333m - length of aircraft carrier deck (Nimitz Class - http://en.wikipedia.org/wiki/Aircraft_carrier)
20,411.6567 - weight of jet in kilos
2 = number of seconds a jet is active on the catapult (assume it clears all 333ms)
165mph takeoff speed = 265.54176 Km/h = 73.7616 m/s
Force = weight * (333/4) - Newton's second law of motion (F = MA | F = kg (m/s^2))

1,699,270.420 Netwons

(1,699,270.420 / 3401.94278) = acceleration of truck = 499.5 m/s^2

Height of unloaded carrier sitting in the water - 30m

Standard gravity - 9.80665 m/s^2

Bust out some calculus:

Position function x(T) = x(0) + v(0)T + (1/2)aT^2
given that acceleration due to gravity is constant
[Position should be 30m at time T - inverted the Y axis so I don't have to deal with imaginary numbers] 
30M = (1/2)(9.80665)T^2
And given that we want to solve for T..
T = SQRT(30/[4.903325])) = 2.4735192091163491620183773821736 seconds until the truck hits the water, assuming there's no lip on the flightdeck

So let's plug that time value into our position function for senor truck...
x(2.4735) = 0 + 0(2.4735) + (0.5)(499.5)(2.4735^2) = 1528.045m

The truck will travel 1528.045m, or roughly 0.949483143 miles.


----- SOLUTION USING JET TAKEOFF VELOCITY -----
v(2) = 73.7616
v(t) = v(0) + aT
a = 73.7616/2 = 36.8088 m / s^2 [way lower than using the length of the take-off deck as the distance traveled in 2 seconds]

F = 20,411.6567 * 36.8088 = 751328.58913896 Newtons of force on the jet

Acceleration on the truck = (751328.58913896 / 3401.94278) = acceleration of truck = 220.8 m/s

Plug back into original position function... 
x(2.4735) = 0 + 0(2.4735) + (0.5)(220.8)(2.4735^2) = 675.4495284m

It is clear to me that the catapult on a Nimitz class carrier does not span the entire length of the flight deck - hence why the first answer is off by an order of magnitude from the second.