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Homework Help: Dennis and Carmen problem with my trial solution , is it right?

  1. Oct 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Dennis and Carmen are standing on the edge of a cliff. Dennis throws a basketball vertically upward, and at the same time Carmen throws a basketball vertically downward with the same initial speed. You are standing below the cliff observing this strange behavior. Whose ball is moving fastest when it hits the ground?

    2. Relevant equations

    v^2 = v0^2 -2gΔy

    3. The attempt at a solution
    in the same speed cause when Dennis basketball back to the initial position "`the throwing position" it will back with the same initial speed with different direction "the same direction as Carmen" so it's the same but it'll take Dennis basketball more time
    v0 is +
    v^2 =+v0^2 -2gΔy
    v0 is -
    v^2 =-v0^2 -2gΔy
    g =-9.8 for both
    Δy is the same for both

    Dennis v^2 =+v0^2
    Carmen v^2 =-v0^2
    add them
    v^2+v^2 =0

    v^2 =-v^2

    please tell me if that's right if not how it can be solve
    thank you:smile:
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 9, 2013 #2


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    Staff: Mentor

    Your logic is correct :approve:

    Careful, the formula wants you to square the initial velocity. Carmen's initial velocity is -v0, and squared is (-v0)(-v0) = +v02.

    As you can see, the initial velocity being negative does not change the result.

    If you consider that last line of math it represents an impossible situation, since you can't have any real number squared that turns out negative. The glitch can be traced back to squaring the initial velocity as I pointed out above.

    You could also have used a conservation of energy approach (if you've covered that yet in your course).
  4. Oct 9, 2013 #3
    thank you ,we have not cover it yet
    so if I'm going to answer for that question in the exam i just write this formula "v^2 = v0^2 -2gΔy" and say
    everything is the same for both cases " v0^2 -2gΔy"
    then the final speed is the same
    only or should i add something else ?
  5. Oct 9, 2013 #4


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    Gold Member

    Forget the math for a minute and just think about it this way. If the ball thrown upward has an initial velocity X, then when it gets back to the same point on its way down, what is its velocity at that point?
  6. Oct 9, 2013 #5


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    Staff: Mentor

    That's all you really need :smile:
  7. Oct 9, 2013 #6
    the same initial velocity that was it thrown with
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