= Free fall in a constant gravitational field
{title2=$s(t)=\tfrac12gt^2,\quad v(t)=gt$}
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 gives
$$
v(t)=gt,\qquad s(t)=\frac12gt^2.
$$
For a downward distance $h>0$, the arrival time and <speed> are
$$
t_{\mathrm{fall}}=\sqrt{\frac{2h}{g}},\qquad v_{\mathrm{arrival}}=\sqrt{2gh}.
$$
For the model values $h=20\,\mathrm m$ and $g=9.81\,\mathrm{m\,s^{-2}}$, these are approximately $2.0193\,\mathrm s$ and $19.8091\,\mathrm{m\,s^{-1}}$. Doubling the height multiplies both quantities by $\sqrt2$; it does not double them. The same <speed> follows from the <work-energy theorem>, $mgh=mv^2/2$.
For intervals of length $\Delta t$, the successive downward <displacements> are
$$
s((j+1)\Delta t)-s(j\Delta t)=\frac12g(\Delta t)^2(2j+1),\qquad j=0,1,2,\ldots.
$$
Thus 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 $h$ is
$$
\frac{g(R+h)}{g_0}=\left(1+\frac hR\right)^{-2}.
$$
For $h/R\ll1$, 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.
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