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Math Help - 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 -g the minus sign is conventional as
    up is normally taken as positive, and so downward accelleration is negative.

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

    Integrating again gives:

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

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

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

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

    or:

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

    Putting this back into the equation for vertical velocity:

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

    At the earth surface 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 his

    PE=m \cdot g \cdot h,

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

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

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