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Object thrown vertically

  1. Sep 24, 2004 #1
    An object is thrown vertically upward such that it has a speed of 58 m/s when it reaches two thirds of it's maximum height above the launch point.
    The acceleration of gravity is 9.8 m/s^2

    Find the maximum height h. Answer in units of m.


    I know:

    V: 58 m/s at y=2/3 H
    H = ?
    V = 0

    V^2-Vo^2 = 2g(1/3H)

    I know I need to plug in the numbers, but how do I get all of the equation to one side, to find H??
     
  2. jcsd
  3. Sep 24, 2004 #2

    Doc Al

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    Staff: Mentor

    Right (where Vo = 0).

    Divide or multiply both sides to eliminate what's in front of H.

    For example, if you had an equation: 7x^2 = 67bZ, and you wanted to isolate Z, just divide both sides by 67b: (7x^2)/67b = (67bZ)/67b = Z; so Z = (7x^2)/67b. Make sense?
     
  4. Sep 24, 2004 #3

    Pyrrhus

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    You got

    [tex] V^2 = V_{o}^2 + 2g\frac{2}{3}H [/tex]

    where V is 58 m/s

    and

    [tex] 0 = V_{o}^2 + 2gH [/tex]
     
    Last edited: Sep 24, 2004
  5. Sep 24, 2004 #4

    Pyrrhus

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    Doc Al, how come initial speed = 0 ?
     
  6. Sep 24, 2004 #5

    Doc Al

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    If you look at the motion on the way down, then the object starts from rest and falls a distance of H/3 to reach a speed of 58 m/s.
     
  7. Sep 24, 2004 #6

    Pyrrhus

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    Oh Yes, i wasn't thinking about it :biggrin:
     
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