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Free fall with horizontal velocity

  1. Feb 18, 2009 #1
    1. The problem statement, all variables and given/known data
    An airplane releases a spacecraft, after 2 seconds the spacecrafts engine starts providing it with a horizontal velocity of 6m/s relative to the aircraft. What is the magnitude and direction of the velocity after 2 seconds? After 3? Assume the spacecraft is in free-fall in the y direction.


    2. Relevant equations
    x=vit+1/2at^2
    y=-1/2gt^2

    3. The attempt at a solution
    It seemed pretty easy... Plug t=2 into the y equation and get a y displacement of 19.6 meters. Then plug in t=3 and arrive at a y displacement of 44.2m, and an x displacement of 3m. Then take the arctan(44.19/3) to get an angle of -86 degrees. The problem is that the answer book appears to split the reference frame at t=2. They say that the y displacement between 2 and 3 seconds is only 4.91m. This doesn't make sense because the object should continue to accelerate downward regardless of the change in x acceleration. I would just assume that this was an error in the answer book, but that would be 3 in a row, and I don't trust myself that much. Any help would be appreciated.
     
  2. jcsd
  3. Feb 18, 2009 #2

    LowlyPion

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    Don't you want instantaneous velocity?

    After 2 seconds won't the dropping V be -19.6 m/s

    After 3 horizontal will be 3 and vertical -g*t = -29.4 m/s

    And overall |V| will be (32 +29.42)1/2 and direction via arc tan?
     
  4. Feb 19, 2009 #3
    I know I'm nit-picking, but doesn't the problem state that the spacecraft is falling in the y direction. So those speed should be positive.
     
  5. Feb 19, 2009 #4

    LowlyPion

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    It doesn't really matter so long as the direction chosen as positive is maintained through out.

    I merely continued to use the direction already indicated by the OP's equation for y.

    In general my preference is usually positive up anyway.
     
  6. Feb 20, 2009 #5
    nah, looking for displacement, and your answer doesn't match their's anyway. They still get a total displacement of 19.6 meters to t=2. But looking at it more, they ignore the acceleration of the object over the first two seconds when calculating displacement over (2,3)
     
  7. Feb 20, 2009 #6

    LowlyPion

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    Are you looking for displacement or velocity? The original problem says velocity doesn't it?

    As to my equation I see that I took the horizontal V to be 3 and in rereading the problem I see that should have been 6m/s at 3s.
     
  8. Feb 20, 2009 #7
    doh... i miscopied the problem... that should say displacement after 2 seconds, sorry about that. Anyway, I talked to the professor, it was an answer book error, thanks for trying anyway.
     
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