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Conservation of Energy

  1. Oct 12, 2008 #1
    A stone is thrown upward at an angle of 53 degrees above the horizontal. Its maximum height above the release point is 24m. What was the stone's initial speed?
    Assume any effects of air resistance are negligible.

    This problem is to be solved using Conservation of Energy.

    Since there is no external force nor non-conservative forces present, this is just the change in mechanical energy.

    Uf + Kf = Ui + Ki
    mgh + 0 = 0 + 1/2 mv^2
    =mgh = 1/2 mv^2
    where y = 0 at the horizontal.

    I got stuck because the mass of the stone is not given. Is my approach wrong?
     
  2. jcsd
  3. Oct 12, 2008 #2

    G01

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    Divide both sides of your last line by the mass. Notice that the mass cancels out of the equation! You don't need to know it.
     
  4. Oct 12, 2008 #3
    Doh!!
    Thanks for pointing that out.
     
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