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Kinematics of rolling ball question

  1. Sep 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Hello, new here. I'm doing homework for my Honors Physics class, and I have a kinematics problem that seems simple enough, but I just can't get it. Help, please?

    A ball is rolling on a table 1.225 m high at 6 m/s. After it rolls off, how long does it take to hit the ground?


    2. Relevant equations

    I used V^2= (Vo)^2 + 2a(X-Xo) to find final velocity, than used V= Vo+at to find time.


    3. The attempt at a solution

    V^2= (6 m/s)^2 + 2(9.8 m/s^2)(1.225 m) V= 7.7 m/s

    t= (V-Vo)/a = (7.7 m/s - 6 m/s)/ 9.8 m/s^2 t= .17 s

    I know this is wrong. My physics teacher told me to use Vx, not Vy, for Vo. I'm not quite sure how to go about this. Do I need to find the angle (which I'm not sure how to do in this problem) and use Vox= Vocos(theta)?

    Any help would be appreciated. :redface:
     
  2. jcsd
  3. Sep 8, 2007 #2

    learningphysics

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    Homework Helper

    On the table the ball only had a horizontal velocity... that only affects horizontal displacement... it doesn't affect vertical displacement.

    What is the initial vertical velocity? You have a vertical displacement of -1.225m.
     
  4. Sep 8, 2007 #3
    It doesnt matter what the horizontal velocity is, the 6m/s is there to confuse you. Only consider the vertical motion, Use Distance= Vo + 1/2at^2 where Vo (the initial vertical velocity) is 0. Oh, and only consider the distance travelled to be negative if you consider the acceleration to be negative.
     
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