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One ball dropped, another ball launched

  1. May 12, 2017 #1
    1. The problem statement, all variables and given/known data
    A ball is dropped from rest at height ##h_0## above the ground. At the same instant, a second ball is launched with speed ##v_0## straight up from the ground, at a point directly below where the other ball is dropped. (a) Find a condition on ##v_0## such that the two balls will collide mid-air. (b) Find an expression for the height at which they collide.

    2. Relevant equations


    3. The attempt at a solution

    Attempted solution for (a): I think all that is required is that ##v_0 > 0##. As long as the velocity is positive then the ball that is launched upward will be in the air by the time it collides with the ball that was dropped. But the textbook gives as a solution that we must have ##v_0 > \sqrt{gh_0/2}## where ##g = 9.8 m/s^2##. What's wrong with my thinking?

    Attempted solution for (b): Let ##x_{dropped}(t)## be the position function of the ball that's dropped. Then ##x_{dropped}(t) = \frac{g}{2}t^2 + 0(t) + h_0 = \frac{g}{2}t^2 + h_0##. Now let ##x_{launched}(t)## be the position function of the ball that is launched. Then ##x_{launched}(t) = \frac{g}{2}t^2 + v_0t + 0 = \frac{g}{2}t^2 + v_0t##.

    I can find at what time the two balls collide by equating the two above equations and solving for ##t##:

    ##x_{dropped} = x_{launched}##
    ##h_0 = v_0t##,

    which gives ##t = \frac{h_0}{v_0}##. Plugging this solution into either ##x_{dropped}## or ##x_{launched}## gives the height of the collision as ##\frac{g}{2}\big(\frac{h_0}{v_0}\big)^2##. But the textbook's solution is ##h_0 - \frac{gh_0^2}{2v_0}##.

    Could someone please tell me why my answers are wrong? Thanks.
     
  2. jcsd
  3. May 12, 2017 #2

    haruspex

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    You need them to collide before the lower ball falls to the ground.
     
  4. May 12, 2017 #3

    haruspex

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    No it doesn't. Try those again.
    You appear to have chosen g to be the upward acceleration due to gravity, making it g=-9.8m/s2. I have no problem with that, but most people make g as positive downwards, writing the acceleration as -g, so that g has a positive value.
     
  5. May 12, 2017 #4

    PeroK

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    It's interesting that this equation is independent of ##g##. Why do you think that is?

    In fact, it's just a simple equation for one ball moving towards a stationary ball.
     
  6. May 12, 2017 #5

    I'm on my phone, so please excuse the no LaTeX... So choosing the natural orientation for my coordinate system, I get x_dropped = -gt^2/2 + v_0t and x_launched = -gt^2/2 + h_0. Setting them equal and solving for t gives me t = h_0/v_0. Plugging this into one of my original equations gives -gh_0/(2v_0^2) + h_0. This is very similar to the textbook's answer, but is still not the same. I can't see where I am going wrong.

    @PeroK I am guessing that my solution for t is independent of g because both balls are undergoing the exact same acceleration (the acceleration caused by gravity)
     
  7. May 12, 2017 #6

    PeroK

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    You've simply not substituted ##t = \frac{h_0}{v_0}## correctly.

    Yes, that's correct.
     
  8. May 12, 2017 #7
    Sorry, typo. I meant to type -gh_0^2/(2v_0^2) + h_0, but this is still different than the textbook's answer (their v_0 is not squared in the denominator)
     
  9. May 12, 2017 #8

    PeroK

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    What answer are you talking about?
     
  10. May 12, 2017 #9

    PeroK

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    This textbook answer is clearly wrong.
     
  11. May 12, 2017 #10
    Thanks for the help. I thought their answer may have been wrong too but because I am new to physics I am not comfortable assuming that.
     
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