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Kinematics with one variable

  1. Sep 6, 2008 #1
    1. The problem statement, all variables and given/known data
    A stone is thrown vertically upward. On its way up it passes point A with speed v, and point B, 5.3 m higher than A, with speed v/2. Calculate the maximum height reached by the stone above point B.


    2. Relevant equations
    v(final) = v(initial) + at
    x(final) = x(initial) + v(initial)*t + 1/2 at^2
    x(final) - x(initial) = 1/2 [v(final) - v(initial)]*t
    2a[x(final) - x(initial)] = v(final)^2 - v(initial)^2

    3. The attempt at a solution
    Vb^2 = Va^2 - 2 * g * Xab, where

    Xab is given = 5.3 m
    Va = V
    Vb = V/2

    (V^2) / 4 = V^2 - g * Xab

    3/4 * V^2 = 2 * g * Xab

    V^2 = 8/3 * 9.8 * 5.3 = 138.5 m^2/s^2
    V = 11.77 m/s

    Now use same equation between B-->C:

    Vc^2 = Vb^2 - 2 * g * Xbc, where

    Vc = 0
    Vb = V / 2 = 11.77 / 2 = 5.88 m/s

    solve for Xbc:

    Xbc = Vb^2 / (2 * g)
    Xbc = (5.88)^2 / (2 * 9.8)
    Xbc = 1.76 m

    So the height above point B is just

    H = Xab + Xbc = 5.3 + 1.76 = 7.06 m



    The above answer is not right.......help
     
  2. jcsd
  3. Sep 6, 2008 #2

    Kurdt

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    Staff Emeritus
    Science Advisor
    Gold Member

    I believe that should be the answer since the question asks for the max height beyond point B.
     
  4. Sep 7, 2008 #3
    ^ Thanks.........silly me, didn't paid attention on that part
     
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