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How long after the second ball is thrown do the two balls pass each other?

  1. Jan 21, 2013 #1
    1. The problem statement, all variables and given/known data
    A juggler performs in a room whose ceiling is 3m above the level of his hands. He throws a ball upward so that it just reaches the ceiling. a) What is the initial velocity of the ball? b) What is the time required for the ball to reach the ceiling? At the instant when the first ball is at the ceiling, he throws the second ball upward with two-thirds the initial velocity of the first c) How long after the second ball is thrown did the two balls pass each other? d) What distance above the juggler's hand do they pass each other?

    2. Relevant equations

    Vy=Vo +a*t

    y-yo=(voy+vy)/2 *t

    y=yo+vo*t + 1/2*a*t^(2)


    3. The attempt at a solution

    I got A and B right.

    A)
    y-yo=(voy+vy)/2 *t

    3m=vo/2*t

    t=6m/vo

    vy=vo+a*t

    vo=g*t

    vo=g*6m/vo

    vo^(2)=g*6m

    vo=7.672m/s

    B) v=vo+a*t

    0=7.672m/s-g*t
    t=0.7820 seconds

    C) This is where I am stuck at.

    So it asks this: How long after the second ball is thrown do the two balls pass each other?
    Is this referring to right when one is above the other or when they are equal in height? I am confused on this part. Thanks in advance for the help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: Feb 4, 2013
  2. jcsd
  3. Jan 21, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Re: Juggler

    Same height. In other words, they have the same value for their y-position.
     
  4. Jan 21, 2013 #3
    Re: Juggler

    Thanks!
     
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