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Conservation of energy problem (projectile with air drag)

  1. Nov 1, 2008 #1
    1. The problem statement, all variables and given/known data

    A stone with weight w is thrown vertically upward into the air from ground level with initial speed v0. If a constant force f due to air drag acts on the stone throughout its flight, (a) show that the maximum height reached by the stone is

    [tex] h = \frac{v_0^2}{2g(1+f/w)} [/tex]

    (b) Show that the stone's speed is

    [tex] v = v_0 \left( \frac{w-f}{w+f} \right) ^ {1/2} [/tex]

    just before impact with the ground.

    2. Relevant equations

    work done by an external force = change in energy

    3. The attempt at a solution

    I have no trouble with part (a). I need help with (b)

    W_{air} &= (-f) * (2h) \\
    &= -2fh \\
    &= -2f\frac{v_0^2}{2g(1+f/w)} \\
    & = -2f\frac{v_0^2 w}{2g(w+f)}

    [tex]W_{air} = \Delta E = \Delta K = \frac{1}{2}m(v^2 - v_0^2) = \frac{w}{2g}(v^2 - v_0^2)[/tex]

    We can set the two expressions equal

    -2f\frac{v_0^2 w}{2g(w+f)} &= \frac{w}{2g}(v^2 - v_0^2) \\
    -2f\frac{v_0^2 }{(w+f)} &= v^2 - v_0^2 \\
    v^2 &= v_0^2 + -2f\frac{v_0^2 }{w+f} \\
    &= v_0^2 \left( \frac{1-2f}{w+f} \right)

    But this is wrong. Can anyone tell me where I messed up?
  2. jcsd
  3. Nov 1, 2008 #2


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    Hi azure kitsune! :smile:

    Only the last line is wrong. :cry:
  4. Nov 2, 2008 #3
    Waaahhhh!!! *feels so stupid!!!*

    Thanks tiny-tim. :smile:
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