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Advance physics problem

  1. Aug 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Two students are on a balcony 23.4 m above the street. One student throws a ball, b1, vertically downward at 15.5 m/s. At the same instant, the other student throws a ball, b2, vertically upward at the same speed. The second ball just misses the balcony on the way down.

    (a) What is the difference in time the balls spend in the air?

    (b) What is the velocity of each ball as it strikes the ground?
    velocity for b1
    velocity for b2

    (c) How far apart are the balls 0.480 s after they are thrown?


    2. Relevant equations

    s = ut + 1/2at
    Not to sure

    3. The attempt at a solution

    I am considering g = 10 m / sec^2.

    Motion of the first ball:

    =>23.4 = (15.5)t + 10(t^2)/2
    =>23.4 = 15.5 t + 5(t^2)
    =>(t^2) + 3.1t - 4.68 = 0
    => t = (-3.1 + 5.3) / 2 = 0.6 sec

    Motion of the second ball

    =>23.4 = - (15.5)t + 10(t^2)/2
    =>23.4 = -(15.5 t) + 5(t^2)
    =>(t^2) - 3.1t - 4.68 = 0
    => t = (3.1 + 5.3) / 2 = 4.2 sec

    Difference in time = 3.6 sec
     
  2. jcsd
  3. Aug 30, 2009 #2
    Need some help here
     
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