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General Launch Angle Question

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data
    On a hot summer day, a young girl swings on a rope above the local swimming hole . When she lets go of the rope her initial velocity is 2.10 m/s at an angle of 35.0 degrees above the horizontal. If she is in flight for 0.615 s, how high above the water was she when she let go of the rope?


    2. Relevant equations

    The equation I used to try to solve for the problem was y=(v0sin(theta))t-.5gt2

    3. The attempt at a solution

    I know the angle is 35 degrees, initial velocity is 2.10 m/s and that initial X and Y positions are both zero. I plug my numbers into the equation given above and get 3.15 for y. The answer in the back of the book says 1.07 meters, but I can't figure out how they arrived at this answer.
     
  2. jcsd
  3. Sep 26, 2009 #2

    rl.bhat

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    Show your calculations. I am getting different answer.
     
  4. Sep 26, 2009 #3
    Ok, I think I messed the calculation up a little. Here should be the correct calculation.

    y= (2.10 x sin 35).615 - .5(-9.80).6152 = 2.59 or 3
     
  5. Sep 26, 2009 #4

    rl.bhat

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    When you use the formula
    y = vo*t - 0.5*g*t^2, you assume that y and vo are in the upward direction and g is in the downward direction. So you should not use -g in the equation.
     
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