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i believe the answer is a bit surprising. the length of the runway is actually less on the aircraft carrier than if it were to have been in an airport.
i don't feel like using the calculator, but i think the basic equation is as follows
momentum of airplane before is Pa =MaVa
momentum of airplane/carrier after is Pt=MtVt where t = total and Mt = Mc+Ma
so using energy equations
1/2Ma(Va)^2 = uMagx + 1/2Mt(Vt)^2
where u = coefficent of friction (which by the way is always between 0 and 1)
and g = gravity and x is the distance.
all that is easliy solved.
the interesting part is that if it were an airport instead, the final velocity would be 0, and so its just:
1/2mv^2 = umgx, where x in this airport solution would be larger than the x in the aircraft carrier solution.
it seems weird, but we have to remember that the airplane travels a distance longer than the carrier's length because the airplane pushes the carrier through the water as well.
in both cases, solve for x yourself. i don't have a calculator handy nor do i wish to use the accessories calculator.
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