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Thread: 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 $\displaystyle kv^2 $, where v is the velocity and k is a positive constant. Given that the maximum height is $\displaystyle H = \frac{1}{2k}ln (1+\frac{k(v_0)^2}{g}) $ Show that to double the maximum height, $\displaystyle v_0 $ must be increased by a factor of $\displaystyle (e^{2kH} + 1)^\frac{1}{2} $
    where $\displaystyle v_0 $ is the initial velocity

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    Last edited by differentiate; Aug 24th 2010 at 02:46 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]

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

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

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

    For 2H, the equation (1) becomes

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

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

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

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

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

    For 2H, the equation (1) becomes

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

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