Y1 = (1/2) a t^2 = (1/2) (29.4 m/s^2) (4.00 s)^2 = 472 m
V = a t = (29.4 m/s^2) (4.00 s) = 117.6 m/s
t' = V/g = (117.6 m/s) / (9.8 m/s^2) = 11.9 s
Y2 = V t' - (g/2) t'^2 = (117.6 m/s) (11.9 s) - (9.8 m/s^2) (11.9 s)^2/2 = 845.3 m
Y = Y1 + Y2 = 472 m + 845.3 m = 1317.3 m
A rocket moves upward, starting from rest with an acceleration of 29.4 m/s^2 for 4.00 s. It runs out of fuel at the end of the 4.00 s but does not stop. How high does it rise above the ground?
Would I have to set up two separate equations for this? one for the acceleration due to the fuel and one for acceleration due to gravity? I'm not sure how that would work.
Assume accleration of gravity (9.8 m/s^2) and the actual accleration (29.4 m/s^2) do not change in that short 4.00 s interval.
Use Y1 = (1/2) a t^2 to get the altitude at the end of 4.00 s.
Use V = a t to get the velocity at the end of burnout.
Use t' = V/g to get the additional time that is "coasts" upward after burnout.
Use Y2 = Y1 + V t' - (g/2) t'^2 to get the additional altitude increase during time t'.
The total altitude that nit rises is Y = Y1 + Y2
In an actual launch lasting a minute or so, both a and g change, because the rocket loses mass and gets farther from the center of the Earth. They have simplified the problem somewhat.
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