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Trying to find height

  1. Jun 8, 2009 #1
    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 2.05 m/s at an angle of 35.0 degrees above the horizontal.
    If she is in flight for 0.614 , how high above the water was she when she let go of the rope?

    so I think it forms a right triangle and from there I just use the pythagorean theorem and then the Y is my height?

    I just have no idea
     
  2. jcsd
  3. Jun 8, 2009 #2

    Cyosis

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    This is a resistanceless projectile problem, that means her trajectory will be a parabola. You will have to use kinematic equations. What's the kinematic equation for the y-direction motion? What acceleration is she subjected to in the y-direction?
     
  4. Jun 8, 2009 #3
    are you talking about y=Yo+VyoT+1/2Ayt^2?
     
  5. Jun 8, 2009 #4

    Cyosis

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    I am indeed. You want to find [itex]y_0[/itex] in that equation.
     
  6. Jun 8, 2009 #5
    I think acceleration is 9.81
     
  7. Jun 8, 2009 #6

    Cyosis

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    Almost, don't forget gravity is pulling her down while she is flying up so it's -9.81. You can also calculate v_y and you know t.
     
  8. Jun 8, 2009 #7
    Isn't y=0 and Vyo 2.05 m/s ?
     
  9. Jun 8, 2009 #8

    Cyosis

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    She stops being in flight when she hits the surface of the water, which is where we have set our 0 point, so indeed y=0. However vy0 is not 2.05 m/s v=2.05 m/s, which is a vector that makes an angle of 35 degrees with the horizontal. You want to know the y-component of this vector.
     
  10. Jun 8, 2009 #9
    Oh OKay! I see what you're saying so in order to get Voy you have to use 2.05 in the equation 2.05sin(35) and then plug in the equation from there! That makes sense! Thank you so much! But, I have a question, I thought that you only used that equation for magnitude like y=r * sin() ....you can place velocity in there too ?
     
  11. Jun 8, 2009 #10

    Cyosis

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    Yep that is correct. I am not entirely sure what your last question is. However you always use the equation along the path the acceleration is in. The only acceleration in this problem is g, which acts along the y-direction. You also know that when she lets go she first goes up and then goes down. Therefore her trajectory is a parabola with a maximum, instead of minimum. So the constant in front of the quadratic term needs to be negative.
     
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