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Angle of falling object

  1. Dec 18, 2009 #1
    1. The problem statement, all variables and given/known data

    A stone is thrown off a high cliff with speed 30m/s at an angle 0 degrees. At what angle is the stone's velocity 2 seconds after it's thrown. (measired with respect to the horizontal x direction(angle will be negative))

    2. Relevant equations

    angle = tan(Vy/Vx)



    3. The attempt at a solution
    V = velocity
    Vx = stays constant = 30m/s
    Vy = Vyo + ay*t = 30 m/s + 19.6 m/s = 49.6 m/s

    angle = tan(49.6/30) which is wrong
     
  2. jcsd
  3. Dec 18, 2009 #2

    Doc Al

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    Staff: Mentor

    What's Vyo equal to?
     
  4. Dec 18, 2009 #3
    Then I guess it would be 0 m/s?

    If it was 0 then I would have Vy = 19.6 and then

    tan(19.6/30) is still not the right answer
     
  5. Dec 18, 2009 #4

    Doc Al

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    Seems right to me. Don't forget that Vy and thus the angle are both negative.
     
  6. Dec 18, 2009 #5
    tan((-19.6) / 30) = -0.765477483

    The correct answer(supposedly) is -33.2 degrees with respect to the horizontal +x direction.
     
  7. Dec 18, 2009 #6

    Doc Al

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    That should be inverse tan, not tan. tanθ = (-19.6/30) → θ = tan-1(-19.6/30). (I misread what you wrote before. :uhh:)
     
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