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Acrobats trajectory with a swing

  1. Oct 13, 2011 #1

    nag

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    1. The problem statement, all variables and given/known data

    Two acrobats, each of 72.0 kg, launch themselves together from a swing, holding hands. Their velocity at launch was 6.80 m/s and the angle of their initial velocity relative to the horizontal was 32.0 degrees upwards. At the top of their trajectory, their act calls for them to push against each in such a way that one of them becomes stationary in mid-air and then falls to the safety net while the other speeds away, to reach another swing at the same height at that of the swing they left. What should be the distance (in m) between the two swings?

    2. Relevant equations



    3. The attempt at a solution

    I can calculate the maximum height they reached, but what I dint understand is when they push each other ...

    thanks for the help
     
  2. jcsd
  3. Oct 13, 2011 #2
    The question does seem a little bit vague but I think I understand. At the maximum height there is only velocity in the horizontal direction. So when acrobat A pushes off of acrobat B, acrobat B loses all of his/her horizontal velocity and falls straight downwards while acrobat A gets a nice boost to the next swing. You will basically need to find out the force that acrobat A needs to apply to acrobat B to make acrobat B fall straight downwards. Then use that same force to find out how much farther acrobat B will go.
     
  4. Oct 17, 2011 #3
    if you haven't found the answer yet, steps to follow:

    separate V into Vx and Vy

    for the vertical component
    use (Vy2-Vy1)/a=t to figure out t
    this is the time it takes to get from swing 1 to max height

    for horizontal component
    d=Vxt find d
    (note the t is the same as the t you calculated above)
    this is the horizontal distance from swing one to when max height occurs

    now use momentum equation. specifically:

    Vx(m1+m2)=m1v1
    since m1=m2

    Vx(2m)=mv1
    v1=2Vx

    use the horizontal component equation of

    D=Vxt to find D (again).
    this time, Vx=v1 (horizontal velocity has changed) so d=v1t

    but if you realize this time, the D is horizontal distance from when max height occurs to the second swing.

    to answer your question, add d and D.

    tada.

    good luck with your CAPA
     
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