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Quick question on a one-dimesnional motion problem

  1. (a): at the halfway point

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  2. (b): above the halfway point

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  3. (c): below the halfway point

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  1. Sep 11, 2014 #1

    Below is the problem listed directly out of the book:

    "you drop a ball from a window located on an upper floor of a building. It strikes the ground with speed v. You now repeat the drop, but your friend down on the ground throws another ball upward at the same speed v, releasing her ball at the same moment that you drop yours from the window. At some location, the balls pass each other. Is this location (a) at the halfway point between the window and the ground, (b) above this point, or (c) below this point.

    *Our teacher later clarified that one should consider the results as if the ball leaves the ground at speed v (no acceleration, and not having taken off from 'your friend's hand' somewhere above the ground).

    2. Relevant equations

    Since there are no hard numbers attached, the only one I can think of is a=9.8m/s^2

    3. The attempt at a solution

    So, when I drew out a graph (I used standard gravitational acceleration and a four second interval), I saw that at at the halfway point in time the ball thrown from the roof was significantly above the halfway point and the one thrown from the ground was significantly above this point. Beyond that, one can reason that due to the acceleration of the ball, the 'fast half' of the journey of each ball is from this midpoint to the ground. After checking this reasoning against many of my peers, as well as my calculus teacher, they confirmed I'm right.

    However, the book, as well as my physics teacher claims that this problem when graphed, forms two inverse, completely linear acceleration graphs, and therefore the balls meet at the midpoint between the ground and the roof.

    This can't be, can it? I believe my reasoning is sound, but I'm just a first year physics student arguing with my AP physics teacher who is sure she is right...

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Sep 11, 2014 #2


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    Gold Member

    Have you tried running through the calculations rather than graphing?

    You know the acceleration of both balls, and you know the initial speed and position of both balls.

    Quick question about you're calculus experience (since you mentioned AP). Are you just doing derivatives or are you doing integrals (~antiderivatives) as well?
  4. Sep 11, 2014 #3
    Your Physics teacher is mistaken. They meet 3/4 of the way up.

    At that point in time, their velocities are -1/2v and 1/2v.

    Edit: If you are up for it, I recommend that you find the general case for d1(t) and d2(2) at tf, where tf (t final) is the time the ball that is dropped hits the ground. Evaluate both at tf/2, and show that d1(tf/2) = d2(tf/2). Then take the result and compare it to 1/2*d1(0).
    Last edited: Sep 11, 2014

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