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Thread: velocity of falling object

  1. #1
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    velocity of falling object

    If you drop a 1.36 kg weight and it falls .1524 meters, what is its velocity when it hits the ground and how is this calculated. Ultimately I am trying to decide how much energy is released.

    Thanks,
    Greg
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Greg E
    If you drop a 1.36 kg weight and it falls .1524 meters, what is its velocity when it hits the ground and how is this calculated. Ultimately I am trying to decide how much energy is released.

    Thanks,
    Greg
    An object dropped near the surface of the earth experiences an approximately
    constant acceleration of $\displaystyle -g$ the minus sign is conventional as
    up is normally taken as positive, and so downward accelleration is negative.

    So $\displaystyle v=-g\cdot t+c$, and if the initial speed is $\displaystyle 0$ then $\displaystyle c=0$.

    Integrating again gives:

    $\displaystyle x=-\frac{g\cdot t^2}{2}+x_0$,

    where $\displaystyle x_0$ is the initial height of the weight, for convenience we will
    take this to be $\displaystyle 0$.

    Now we are interested in what is happening when the weight has fallen
    $\displaystyle 0.1524m$. This occurs when:

    $\displaystyle 0.1524=-\frac{g\cdot t^2}{2}$,

    or:

    $\displaystyle t=\sqrt{\frac{2\cdot 0.1524}{g}}$.

    Putting this back into the equation for vertical velocity:

    $\displaystyle v=-g \cdot \left( \sqrt {\frac{2 \cdot 0.1524}{g}} \right)=-\sqrt{g \cdot 2 \cdot 0.1524}$.

    At the earth surface $\displaystyle g \sim 9.81 m/s$.

    We could have short circuited this calculation by knowing that the change
    in potential energy when the weight drops through a height$\displaystyle h$is

    $\displaystyle PE=m \cdot g \cdot h$,

    which will equal the kinetic energy of the weight when it has fallen from rest
    through a distance $\displaystyle h$:

    $\displaystyle KE= \frac{1}{2}m \cdot v^2$

    RonL
    Last edited by CaptainBlack; Dec 21st 2005 at 12:53 PM.
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