How to obtain the distance traversed by a free falling body equation? I know that the distance that a free falling body has traverse through time is given by $d=0.5*g*t^2$. I would like to know how to get to this equation to study a bit more how it was obtained. I have been searching around with any luck. Could anybody give any tip?
 A: The equation comes from Newton's second law:
$$ F = ma $$
Galileo didn't know calculus (because Newton and Leibniz hadn't discovered it yet) so he couldn't derive the equation mathematically. Since we do know calculus we know that acceleration is the variation of velocity with time:
$$ a = \frac{dv}{dt} $$
And also the gravitational force $F$ is equal to $mg$. If we substitute for $a$ and $F$ in Newton's second law we get (after a slight rearrangement):
$$ \frac{dv}{dt} = g $$
Knowing calculus we can integrate both sides to get:
$$ v = gt + C $$
where $C$ is the constant of integration. To find $C$ we note that when $t = 0$ we find $v = C$, so $C$ is the velocity when we start timing. In the particular case of dropping an object the initial velocity is zero, but for now let's keep the initial velocity and call it $u$, which is the usual symbol for it, and our equation becomes:
$$ v = u + gt $$
The next step is to note that the velocity $v$ is the variation of distance with time:
$$ v = \frac{ds}{dt} = u + gt $$
and we can integrate this to get:
$$ s = ut + \tfrac{1}{2}gt^2 + D $$
where once again $D$ is a constant of integration. This time we note that when $t = 0$ the distance $s = 0$, so the constant $D$ is zero and our final equation is:
$$ s = ut + \tfrac{1}{2}gt^2 $$
The equation you give the the special case of an object starting at rest, i.e. $u = 0$, in which case the equation simplifies to:
$$ s = \tfrac{1}{2}gt^2 $$
There are a number of related equations derived in a similar way, and they are generically referred to as the SUVAT equations.
