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Homework Help: Simple kinematics ball problem

  1. Jan 16, 2010 #1
    1. The problem statement, all variables and given/known data

    A ball is thrown straight upward with an initial speed v0. When it reaches
    the top of its flight at height h, a second ball is thrown straight upward with
    the same initial speed. Do the balls cross paths (a) at height 1⁄2h (b) above
    1⁄2h or (c) below 1⁄2h. Explain your reasoning.


    3. The attempt at a solution

    My approach was quite simplistic. Since the second ball has an initial speed, over the given time interval, it'll move over a longer distance than the first one. That was also what the answers said. However, I don't find this answer satisfying. I would like to prove this analytically but unfortunately I can't find a way to do so.

    I tried to write out the distance equations for both balls:

    [tex]
    h=v_0t - \frac{gt^2}{2}
    [/tex]
    [tex]
    h=h_t - \frac{gt^2}{2}
    [/tex]

    where [tex]h[/tex] is the height at which the balls cross paths, [tex]h_t[/tex] is the max height the first ball reaches and [tex]v_0[/tex] is the initial speed. Now the RHS's of both equations are equal, so equating them and solving for t gives

    [tex]
    t=\frac{h_t}{v_0}
    [/tex]

    Substituting that back into the first of the two equations gives

    [tex]
    h=h_t-\frac{gh_t^2}{2v_0^2}
    [/tex]

    From here, I can't get any further. I'm not even sure this is the right approach.

    Thank you for any help.
     
  2. jcsd
  3. Jan 16, 2010 #2
    Right, I'm a bit too lazy to mull over your working, so I might be kinda starting off almost anew.

    First thing, the maximum height that the ball reaches can be easily represented in terms of other variables. So,
    [tex]h_{max} = \frac{v_{0}^{2}}{2g}[/tex]​

    Now, let the balls meet after a time t. Then, the distances travelled by ball 1 and ball 2 respectively are
    [tex]d_{1} = \frac{1}{2}gt^{2}[/tex]
    [tex]d_{2} = v_{0}t - \frac{1}{2}gt^{2}[/tex]

    We further know that the total distance must be equal to the max height attained by ball 1.
    [tex]d_{1} + d_{2} = v_{0}t = h_{max} = \frac{v_{0}^{2}}{2g}[/tex]​
    Then,
    [tex]t = \frac{v_{0}}{2g}[/tex]​
    Which gives
    [tex]d_{2} - d_{1} = v_{0}t - gt^{2} = \frac{v_{0}^{2}}{2g} - \frac{v_{0}^{2}}{4g} = \frac{v_{0}^{2}}{4g} > 0[/tex]
    Hence, we are done in showing that d2 is greater than d1.
     
    Last edited by a moderator: Sep 28, 2016
  4. Jan 16, 2010 #3
    Oh, that makes sense.

    Cheers mate.
     
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