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Math Help - Applications of Calculus - maximum height

  1. #1
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    Thumbs up Applications of Calculus - maximum height

    A body is projected vertically upwards from the ground in a medium that produces a resistance force per unit mass of  kv^2 , where v is the velocity and k is a positive constant. Given that the maximum height is   H = \frac{1}{2k}ln (1+\frac{k(v_0)^2}{g}) Show that to double the maximum height,  v_0 must be increased by a factor of  (e^{2kH} + 1)^\frac{1}{2}
    where  v_0 is the initial velocity

    Thanks
    Last edited by differentiate; August 24th 2010 at 03:46 AM.
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  2. #2
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    [ΜΑΤΗ]Η = \frac{1}{2k}ln(1 + \frac{k(v_o)^2}{g}[/tex]

    2kH = ln(1 + \frac{k(v_o)^2}{g}

    e^{2kH} = (1 + \frac{k(v_o)^2}{g}

    e^{2kH} - 1 =  \frac{k(v_o)^2}{g} .......(1)

    For 2H, the equation (1) becomes

    e^{4kH} - 1 =  \frac{k(v_o')^2}{g} ........(2)

    Now divide (2) by (1) and simplify to find vo'
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  3. #3
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    H = \frac{1}{2k}ln (1+\frac{k(v_0)^2}{g})

    2kH = ln(1 + \frac{k(v_o)^2}{g})

    e^{2kH} = (1 + \frac{k(v_o)^2}{g})

    e^{2kH} - 1 =  \frac{k(v_o)^2}{g} .......(1)

    For 2H, the equation (1) becomes

    e^{4kH} - 1 =  \frac{k(v_o')^2}{g} ........(2)

    Now divide (2) by (1) and simplify to find vo'
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