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Test tomorrow, can't figure this problem out

  1. Feb 1, 2006 #1
    An object is thrown vertically upward such that it has a speed of 21 m/s when it reaches two thirds of its maximum height above the launch point. Determine this maximum height.

    I can't figure what to do since initial velocity isn't given and I don't see a way to solve for it.
  2. jcsd
  3. Feb 1, 2006 #2

    Doc Al

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    Staff: Mentor

    Reframe the problem this way: An object is thrown upward with an initial speed of 21 m/s and rises to a height of h/3. Solve for h.
  4. Feb 1, 2006 #3
    Oh I like that suggestion Doc Al. Very clever.

    I would use a kinematic equation to help get your answer... this one perhaps?

  5. Feb 1, 2006 #4
    That equation doesn't really help me though, Jameson. It has two variables in it that I don't know the value of.

    And I kind of see where you're going, Doc Al, but why h/3? It seems like 2h/3 would be what I'd want to solve for...
  6. Feb 2, 2006 #5


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    What doc al is saying is you can find the maximum heigth because at the point [itex]\frac{2}{3}h[/itex] the initial velocity would be 21ms/2 and at maximum height (which is [itex]\frac{1}{3}h[/itex] above that point) you know the ball would come to rest. This would allow you to use Jameson's equation to calculate the maximum height.
  7. Feb 2, 2006 #6

    Doc Al

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    Hootenanny already explained it, but let me put it this way. You have three points of interest:
    (1) start: y = 0; v = ?
    (2) "mid" point: y = 2h/3; v = 21 m/s
    (3) top: y = h; v = 0 m/s

    What I'm suggesting is that between points 2 & 3 the object travels a distance of h/3. A perfect opportunity to apply Jameson's kinematic equation.
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