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Homework Help: Another kinematics problem solved at the last second

  1. Aug 17, 2004 #1
    I am [make that: WAS] having difficulties solving Problem 87 from Chapter 2 of Physics for Scientists and Engineers by Paul A. Tipler, 4th edition. The statement of the problem is as follows:

    Ball A is dropped from the top of a building at the same instant that ball B is thrown vertically upward from the ground. When the balls collide, they are moving in opposite directions, and the speed of A is twice the speed of B. At what fraction of the height of the building does the collision occur?

    My incomplete [<Ahem> Make that: COMPLETE] attempt at a solution goes as follows:

    We denote the height of the building as h, the position of ball A as yA, the initial position of ball A as yA0, and analogously for ball B.

    The equation of motion for ball A is the following:
    yA - yA0 = vA0t - (0.5)gt2
    yA0 = h
    vA0 = 0
    g = 9.81 m/s2
    Therefore, yA = -(0.5)(9.81 m/s2)t2 + h
    => yA = -4.905t2 + h

    The equation of motion for ball B is the following:
    yB - yB0 = vB0t - (0.5)gt2
    yB0 = 0
    g = 9.81 m/s2
    Therefore, yB = vB0t - (0.5)(9.81 m/s2)t2
    => yB = vB0t - 4.905t2

    The two balls collide when yA = yB:

    yA = yB => -4.905t2 + h = vB0t - 4.905t2
    => h = vB0t

    For ball A:
    vA = vA0 - gt
    vA0 = 0
    Therefore, vA = - gt => vA = -9.81t

    For ball B:
    vB = vB0 - gt
    Therefore, vB = vB0 - gt => vB = vB0 - 9.81t

    We also know that when yA = yB, vA = -2vB:

    vA = -2vB => -9.81t = -2[vB0 - 9.81t]
    => -9.81t = -2vB0 + 19.62t
    => 2vB0 = 29.43t
    => vB0 = 14.715t

    Since h = vB0t, we have:

    h = (14.715t)t = 14.715t2

    Unfortunately, I couldn't figure out what to do from this point until I spent the better part of the last hour typing this thread. :grumpy: Then, as I was typing the previous sentence, it hit me:

    xA = -4.905t2 + h = -4.905t2 + 14.715t2
    => xA = 9.81t2

    Therefore, xA / h = 9.81t2 / (14.715t2)
    => xA / h = 2/3
    => xA = 2h/3
    , which is the answer given in the back of the book.

    I've decided to post this thread anyway, partly because someone else might benefit from seeing it, and partly because I spent nearly an hour typing it. :cry:

    This also reminds me of a post by Clausius2 in a previous thread I had posted:

    I don't think this would have worked the last time, but it certainly worked here! :rofl: :biggrin:

    Oh brother ... Good night, everyone! :zzz:
  2. jcsd
  3. Aug 18, 2004 #2
    Nice work, who says maths isn't rewarding?
  4. Aug 23, 2004 #3


    User Avatar

    That would be me. :wink:
  5. Oct 2, 2011 #4
    Hey, I know this threads yonks old, but just wondering, does anyone know where the poster got the very final bit from:

    Thanks :)

    Edit: Isn't it strange how you see it once you've posted the question haha! Thanks anyway, problem solved.
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