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Projectile Motion wind acceleration

  1. Mar 11, 2009 #1
    1. The problem statement, all variables and given/known data
    A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a = (6i + 7j)m/s^2 . At time t = 0, the velocity is (3i)m/s . What are the (a) magnitude and (b) angle of its velocity when it has been displaced by 12.0 m parallel to the x axis?


    i= acceleration in x component , j = acceleration in y component


    3. The attempt at a solution





    vx^2 = (3m/s)^2 + 2 (6m/s^2) (12m/s)


    vx^2= 9m/s + 144m/s

    vx = 12.36


    I did that and got velocity wrong what do I do?
     
  2. jcsd
  3. Mar 11, 2009 #2

    LowlyPion

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    You're only half way home.

    What's the Vy?

    Just because it has no initial velocity in j doesn't mean that it isn't swept by the wind in the j direction.
     
  4. Mar 11, 2009 #3
    So once I find Vy I use Vx^2 + Vy^2 = V^2?????
     
  5. Mar 11, 2009 #4

    LowlyPion

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    That would be the way to do it.

    And tan-1(Vy/Vx) would give the angle to positive x.
     
  6. Mar 12, 2009 #5
    how do I find vy?
     
  7. Mar 12, 2009 #6

    LowlyPion

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    Along Vx how long did it take to cover the 12m?

    Use that time to determine Vy.

    Vy = a*t
     
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