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Maximum elevation

  1. Oct 13, 2005 #1
    This problem seems simple but I'm just not getting the correct answer, please help thank you.

    A test rocket is fired straigt up from rest with a net accerleration of 20m/s^2. After 4 seconds the motor turns off, but the rocket continues to coast upward. What maximum elevation does the rocket reach?

    This is what I get

    X_0=0, a=20m/s^2, t=4 s

    1/2(20m/s^2)(4 s)^2=160m (the correct answer is 487m :cry: )
     
    Last edited: Oct 13, 2005
  2. jcsd
  3. Oct 13, 2005 #2

    hotvette

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    Homework Helper

    You need to think of it as 2 separate problems.

    1. Rocket under the stated acceleration

    2. Rocket after acceleration stops.

    The final conditions of problem 1 (i.e. height and velocity) are the initial conditions for problem 2.
     
  4. Oct 13, 2005 #3
    Problem 1
    x=160m and v=80m/s

    Problem 2
    x_0=160m, v_0=80m/s, a=?
    if it stopped accelerating, I would assume that the acceleration would be decreasing as the rocket coasted upward. Would this be correct?
     
  5. Oct 13, 2005 #4

    hotvette

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    So far, so good. You have the initial conditions for problem #2. Now, you need the equation of motion (in the vertical direction) for a projectile with initial conditions of distance and velocity. You must have been given such an equation or taught how to derive it using F=ma.
     
    Last edited: Oct 13, 2005
  6. Oct 14, 2005 #5
    Thank you. This is what I got

    160m-[0-(80m/s)^2/2(9.8m/s^2)]=486.5m

    Also, I had another question about horizontal projectile motion.
    Why does the distance of two packages steadily increase as they fall out of a plane going at constant speed.

    I thought the distance would remain constant between the two packages as they fall when air friction is negligible. I can't make sense of it.
     
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