# Math Help - Acceleration of Gravity

1. ## 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?

2. Originally Posted by shilz222
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:

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

Therefore:

$
\frac{g^2T_a^2}{4}=v^2+2g(h_0-h_a)
$
.

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

$
\frac{g^2T_a^2}{4}-\frac{g^2T_b^2}{4} = 2g(h_a-h_b)
$

or:

$
\frac{g}{4}(T_a^2-T_b^2)=2(h_a-h_b)
$

and the rest is algebra.

RonL