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Baseball throws

  1. Oct 22, 2011 #1
    1. The problem statement, all variables and given/known data
    x0vg3k.jpg

    2. Relevant equations
    Energy conservation et al.

    3. The attempt at a solution
    I think that all of the final velocities will be equal, but I am not sure how to show this mathematically. Seems like a trick question.
     
  2. jcsd
  3. Oct 22, 2011 #2
    [tex]mgh_{i} + \frac{1}{2}mv_{0}^{2} = mgh_{f} + \frac{1}{2}mv_{f}^{2}[/tex]
    [tex]gh_{i} + \frac{1}{2}v_{0}^{2} = gh_{f} + \frac{1}{2}v_{f}^{2}[/tex]
    Final potential energy is zero at the ground, so:
    [tex]gh_{i} + \frac{1}{2}v_{0}^{2} = \frac{1}{2}v_{f}^{2}[/tex]
    Which gives that:
    [tex]gh_{i} = \frac{1}{2}v_{f}^{2} - \frac{1}{2}v_{0}^{2}[/tex]
     
  4. Oct 22, 2011 #3
    I see that it turns into a familiar kinematic equation. I end up with:

    [tex]v_{f}= \sqrt{2gh+v_{i}^{2}}[/tex]

    Since the initial velocity of all the balls is the same, gravity is the same, and the displacement, well height, is the same, wouldn't that give me that the final velocity is equal to a constant in all cases, and thus they are all equal? Is this correct thinking?
     
  5. Oct 22, 2011 #4
    I put them as all the same, and it's incorrect? What mistake am I making here?
     
  6. Oct 23, 2011 #5

    lewando

    User Avatar
    Gold Member

    Your thinking is correct--they are all the same. Your automated answer submission machine is having a fit.
     
  7. Oct 23, 2011 #6
    It figures. I sent my professor an email on the question. Masteringphysics is so annoying. Thanks for the help.
     
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