 Free fall in a constant gravitational field

Take downward displacement $s$ as positive, assume a constant gravitational acceleration $g>0$, and release a point mass from rest at $s=0$. Integrating $\ddot{s}=g$ twice givesFor a downward distance $h>0$, the arrival time and speed areFor the model values $h=20\mathrm{m}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, these are approximately $2.0193\mathrm{s}$ and $19.8091\mathrm{m}{\mathrm{s}}^{-1}$. Doubling the height multiplies both quantities by $\sqrt{2}$; it does not double them. The same speed follows from the work-energy theorem, $mgh=m{v}^2/2$.

For intervals of length $\Delta t$, the successive downward displacements areThus equal spacing in time produces increasingly separated positions. This is the odd-number pattern in Galileo's result. It assumes constant acceleration and release from rest; a nonzero initial velocity adds a linear term to $s(t)$.

The approximation can be estimated for a spherical source of radius $R$ and standard gravitational parameter $\mu$. Writing $g_0=\mu /R^2$, the gravitational magnitude at height $h$ isFor $h/R\ll 1$, a Taylor expansion gives $g(R+h)/g_0=1-2h/R+O((h/R{)}^2)$. Small height relative to the source's radius makes nearly constant gravity plausible; negligible drag force remains a separate requirement.