Measuring g by free fall

Gist: Drop something from a known height $h$, time the fall, and $g$ falls out of the arithmetic.

Released from rest ($x_0=0$, $v_0=0$) the drop gives $h=\frac{1}{2}g{t}^2$, so

$t=c\sqrt{\frac{h}{g}}c=\sqrt{2}$

Why it matters: Two measurable quantities — one length and one time — yield a constant that cannot be derived from first principles. From $h=3$ m the fall takes about $0.8$ s, and the uncertainty in $g$ follows from the uncertainties in $h$ and $t$.

Detail: Lecture 2, p.6 and p.7

Related: Free fall, Measurement resolution, Stroboscopic measurement, Physics