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Projectile Motion of a juggler

  1. Jun 24, 2008 #1
    1. The problem statement, all variables and given/known data
    A juggler manages to keep five balls in motion, throwing each sequentially up a distance of 3.0 m. Determine the time interval between successive throws (Disregarding the time taken to transfer balls between hands).

    Known variables:
    [tex]g = 9.8 m/s^{2}[/tex]
    [tex]y = 3.0 m[/tex]
    [tex]x_0 = 0[/tex]
    [tex]y_0 = 0[/tex]
    [tex]t = ? [/tex] (looking for)
    [tex]\theta = ?[/tex]
    [tex]v = ?[/tex]


    My main problem is not knowing an angle of elevation (though it must be near 90°) or a velocity. I'm sure there is something simple that I am missing, but I cannot seem to find it.

    2. Relevant equations

    [tex]v_x =v_0cos\theta[/tex] -- initial velocity and angle unknown
    [tex]v_y =v_0sin\theta - gt[/tex] -- initial velocity and angle unknown
    Trajectory: [tex]y = xtan\theta - gx^{2}/(2v_0^{2}cos^{2}\theta)[/tex] -- need initial velocity and angle theta.

    There are others that come to mind, but all seem to have similar problems.

    3. The attempt at a solution

    My initial thought was to find an x value for the maximum, and then use that number to find the time, but then I realized that I did not have the velocity or the angle. I thought of using some of the basic constant acceleration equations, but I constantly ran into problems due to the lack of [tex]t[/tex] and [tex]v_0[/tex]. I think one of my main problems is that I have not been able to incorporate the five balls into my answer as of yet. If somebody could point me in the right direction, it would be great. For those wondering, this is question 27a. in the fourth edition of chapter four of Resnick, Halliday, and Krane's Physics. (And this isn't for school, I'm attempting to teach it to myself. Most of the other problems weren't too bad, but this one is becoming a bit of a nuisance.) Thanks!
     
  2. jcsd
  3. Jun 24, 2008 #2

    dx

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    Gold Member

    It's okay to assume the angle is 90. The velocity can be determined from how high it goes. Think about kinetic and potential energy.
     
  4. Jun 24, 2008 #3
    I tried to assume that for a little while, but then that tangent threw me for a loop there. :P

    A friend pointed me to use [tex]v * v = v_0 * v_0 + 2a * (r - r_0)[/tex], which I hadn't even thought of, and now I think I've figured it out. Thanks for the help, though!
     
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