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Ball dropped/ball thrown

  1. Sep 14, 2008 #1
    1. The problem statement, all variables and given/known data
    You and your friend are on top of a roof.... You simply drop a balloon from rest. Your friend throws a balloon downward 2.7s later with an initial speed of 52.92m/s. They hit the ground at the same time. Neglect air resistance. How high is the apartment house?

    2. The attempt at a solution
    My thinking on this is that after 3.2 seconds balloon 1 has traveled 31.36m, and balloon 2 has traveled 31.36m after .5 seconds, meaning the building would be 31.36m tall, but evidently my thinking is off somewhere.

    So, if anyone could explain what it is I'm doing wrong with this one, it would be much appreciated.

  2. jcsd
  3. Sep 14, 2008 #2


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    Homework Helper

    Hi Michael805,

    I don't believe this is correct. How did you find this result?

    (It appears that you multiplied 9.8 times 3.2, which would give the speed of ball 1 right at t=3.2 seconds, since it has no initial velocity.)

    You do want to find an expression for the displacement (which would be [itex]\Delta x[/itex] or [itex]\Delta y[/itex] in the equation); so what kinematic equation would relate displacement, time, and the quantities they give you in the problem?
  4. Sep 15, 2008 #3
    what formula did you use to get to that answer?
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