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Math Help - Acceleration of Gravity

  1. #1
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    Acceleration of Gravity

    The acceleration of gravity can be measured by projecting a body upward and measuring the time that it takes to pass two given points in both directions.

    Show that if the time the body takes to pass a horizontal line  A in both directions is  T_A , and the time to go by a second line  B in both directions is  T_B , then, assuming that the acceleration is constant, its magnitude is  g = \frac{8h}{T_{a}^{2} - T_{b}^{2}} where  h is the height of line  B above line  A .

    So this is a constant acceleration problem. Also the path is parabolic. How would I deduce the above equality?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by shilz222 View Post
    The acceleration of gravity can be measured by projecting a body upward and measuring the time that it takes to pass two given points in both directions.

    Show that if the time the body takes to pass a horizontal line  A in both directions is  T_A , and the time to go by a second line  B in both directions is  T_B , then, assuming that the acceleration is constant, its magnitude is  g = \frac{8h}{T_{a}^{2} - T_{b}^{2}} where  h is the height of line  B above line  A .

    So this is a constant acceleration problem. Also the path is parabolic. How would I deduce the above equality?
    The height of a projectile as a function of time is:

    x=- ~\frac{gt^2}{2}+v_0t +h_0

    So the projectile passess a height h_a at the roots of:

    - ~\frac{gt^2}{2}+v_0t +(h_0-h_a)=0,

    which are:

    \frac{-v\pm \sqrt{v^2+2g(h_0-h_a)}}{-g}.

    The difference between these roots is:

     <br />
T_a=\frac{2}{g}\sqrt{v^2+2g(h_0-h_a)}<br />
.

    Therefore:

     <br />
\frac{g^2T_a^2}{4}=v^2+2g(h_0-h_a)<br />
.

    Now if we have two such data points we may subtract them:

     <br />
\frac{g^2T_a^2}{4}-\frac{g^2T_b^2}{4} = 2g(h_a-h_b)<br />

    or:

     <br />
\frac{g}{4}(T_a^2-T_b^2)=2(h_a-h_b)<br />

    and the rest is algebra.

    RonL
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