A rocket accelerates by burning its on-board fuel, so its mass decreases with ti

Question

A rocket accelerates by burning its on-board fuel, so its mass decreases with time. Suppose the initial mass of the rocket at liftoff (including its fuel) is m, the fuel is consumed at rate r, and the exhaust gases are ejected with constant velocity v_e (relative to the rocket). A model for the velocity of the rocket at time t is given by the equation where g is the acceleration due to gravity and t is not too large. If g =9.8 m/s^2, m = 35000 kg,r = 160 kg/s, and v_e = 3000 m/s, find the height (in meters) of the rocket one minute after liftoff.

Explanation / Answer

v(t) = -gt - ve(ln((m - rt)/m))

Then:
x(t) = v(t) dt
[-gt - ve(ln((m - rt)/m))] dt

Let u = (m - rt)/m
du = (-r/m) dt
dt = (du)/(-r/m) = (-m/r) du

x(t) = (-1/2)gt^2 - (ve(-m/r)(u(lnu - 1) + C
= ((mve)/r)((m - rt)/m)(ln((m - rt)/m) - 1) - (1/2)gt^2
= (ve(m - rt)/r) (ln((m - rt)/m) - 1) - (1/2)gt^2

g = 9.8 m/s^2
m =35000 kg
r =160 kg/s
ve = 2700 m/s
t = 60 s

x(60) = (35000(35000 - (160*60))/160) ((ln((35000 - (160*60))/35000)) - 1) - ((1/2)(9.8)(60^2))
= (((35000(35000 - 9600)/160) (ln((35000 - 9600)/35000) - 1)) - (17640)

= ((35000(25400))/160)) (ln(25400/35000) - 1) - (17640)

= ((5556250)(1.30)) - (17640)
= 7205485 m

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