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Stuntman jumping 2D kinematics.

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data
    A stuntman is jumping from building A to building B. His flight lasts 1s. He leaves point A with a speed of Vo at an angle of 30 degrees above horizontal. Building B is .9m shorter than building A. What horizontal distance was covered by the stuntman given he makes it to the very edge of building B? What was the maximum height with respect to the starting point attained by the stuntman?

    2. Relevant equations
    Displacement and velocity kinematic equations.

    3. The attempt at a solution
    displacement x= cos(30)*Vo*1s
    displacement y= sin(30)*vo+.5(-9.8)(1)-.9m

    I know that somehow i need to find Vo from the Y data then plug it into the x displacement equation but i am not sure how this is going to be possible with the given information.
  2. jcsd
  3. Sep 8, 2009 #2


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    Your equation giving y has an incorrect value for the initial vertical position. The general form is

    y(t) = y0 + voyt - (1/2)g t2

    What is the value for y0? In other words, if you plug in t=0 in the above equation, what should y0 be so that it matches the vertical position of the stuntman?
  4. Sep 8, 2009 #3
    oh should it be a +.9 since at t=0 he is .9 above where he will end up?
  5. Sep 9, 2009 #4


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    Yes. So his initial vertical position is 0.9 m and his final vertical position is 0 m. Now you can put the equation together.
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