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Initial speed and distance from initial height problem

  1. Jan 11, 2010 #1
    1. The problem statement, all variables and given/known data
    An entertainer juggles balls while doing other activities. In one act, she throws a ball vertically upward, and while it is in the air, she runs to and from a table 5.60 m away at a constant speed of 2.10 m, returning just in time to catch the falling ball.

    2. Relevant equations
    Part A)
    With what minimum initial speed must she throw the ball upward to accomplish this feat?

    Part B)
    How high above its initial position is the ball just as she reaches the table?

    3. The attempt at a solution
    I solved for the first part and got 26.1 by Vf = Vi + at
    Vf = 0
    Vi = 0
    a = 9.8
    t = (11.20/2.10)/2 = 2.67
    9.8*2.67=26.1 m/s
    i don't understand how to do part B of the problem though
  2. jcsd
  3. Jan 11, 2010 #2


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    Homework Helper

    Hi magnifik, welcome to PF.
    You have calculated the initial velocity. You know the time to reach the table. Using the kinematic equation find the height.
  4. Jan 11, 2010 #3
    i'm not exactly sure which equation I use and which time i use
  5. Jan 11, 2010 #4
    I used
    x(t) = vt + (1/2)at^2 + x0
    i plugged in t = 2.67 and v = 26.1
    i got 104.6 from doing this,
    and i'm pretty sure that's wrong because it doesn't seem realistic

    a little guidance would be nice
  6. Jan 11, 2010 #5


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    Homework Helper

    x(t) = vt + (1/2)at^2 + x0
    As the ball goes up, its velocity decreases. So it is retarding. There the equation should be
    x(t) = vt - (1/2)at^2 + x0
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