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A ball is thrown up , another is dropped. Do they arrive at the same t

  1. Apr 24, 2013 #1
    1. The problem statement, all variables and given/known data

    A ball is thrown up , another is dropped. Do they arrive at the same time?

    2. Relevant equations

    3. The attempt at a solution

    I think they wont because the ball which is thrown up has a greater distance to travel, but I am not sure, can someone explain please?
  2. jcsd
  3. Apr 24, 2013 #2


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    Staff: Mentor

    Arrive where?
  4. Apr 24, 2013 #3
    the ground of course..
  5. Apr 24, 2013 #4


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    If I get the question right, you should do the following;

    Assume that the current height of the balls is x. The first one is thrown up whereas the second is dropped. Distance the second ball travels can be found by using the following equation ; x=x0 + V0.t + 1/2at²

    Now, let's say the first ball will go h higher hence the total distance travels is 2h+x. Now, you can do the same above equation and can find the total time.

    Hope, I'm not mistaken.
  6. Apr 24, 2013 #5


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    Hey, what the, just get two balls and perform the experiment.
  7. Apr 25, 2013 #6


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    There is an easy way to look at the problem. The ball that is thrown upwards will eventually come to a halt at some height H. So in effect you have two balls that are being dropped from different heights...and one has a head start.
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