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Homework Help: The velocity and height of a rocket launched vertically from the earth.

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    So if a rocket is launched vertically from the surface of the earth, the rocket has a mass of 1000 kg and has a fuel load of 12000 kg. The fuel burns at such a rate that it will be exhausted after 180 secs. The exhaust velocity of the burned fuel relative to the rocket is 2000 m/sec. Assume an air resistance proportional to the velocity with proportionality constant b = 100 n-sec/m and that the gravity is constant. Find the velocity and height of the rocket when the fuel runs out.

    2. Relevant equations
    Force = mass * acceleration = mass * gravity - air resistance * velocity\
    Mass = 13000 (weight of rocket and fuel) - 66.66t (weight of fuel lost per second or 12000/180)

    3. The attempt at a solution
    So, naturally, I want to find the velocity with the above equation. So with inserting the known values into the equation it becomes,
    (13,000 - 66.66t)a = (13,000 - 66.66t) * 9.8 (gravitational constant - 100 (air resistance) * v
    dividing mass to the other side I get.
    a = 9.8 - (100 * v) / (13,000 - 66.66t)
    Since I will have to integrate a in get it in terms of v, I divided the right half to the left half
    dv/dt (acceleration) / 9.8 - (100 * v) / (13,000 - 66.66t) = 1 dt
    Integrate both sides (I had to use a program to solve this)
    (-6666t - 130)ln(-653.268t - 100v + 127,400) = t
    now we have to get v by itself so,
    ln (-653.268t - 100v + 127,400) = t / (-6666t - 130)
    Get rid of the ln,
    -653.269t - 100v + 127,400 = e ^ (t / (-6666t - 130)
    more simple rearranging gets
    -100v = e ^ (t / (-6666t - 130) + 653.269t - 127,400
    v = -(e ^ (t / (-6666t - 130) + 653.269t - 127,400) / 100
    now we have to subtract the exhaust velocity from all this equaling

    v = (-(e ^ (t / (-6666t - 130) + 653.269t - 127,400) / 100) - 2000

    This should give the velocity of the rocket when the fuel runs out. I theorize that to find the height you simply have to take the integral of this function, and since I don't want to re-write that mess (at least not without knowing if I am actually correct in that assumption), I would greatly appreciate if someone could analyze this work and see if I both found the correct velocity and am correct in how to find the height. Thanks in advance for your help.
  2. jcsd
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