Most analysis of free-fall assume that bodies fall with constant acceleration. If however one analyses free-fall according to Newton's gravitation law, one is lead to a differential equation which I don't know how to solve.
The differential equation one is lead to[1] is $\frac{d^2x}{dt^2}=-\frac{1}{x^2}$. Does this equation have a closed form solution with given initial conditions $x_0,v_0$?
[1] If one picks the units so that $MG=1$, fixes one mass at $x=0$ and places the falling mass at some $x>0$.
Best Answer
Multiply the equation by $dx/dt$ and integrate once to get $$ \frac{1}{2}\Bigl(\frac{dx}{dt}\Bigr)^2=\frac{1}{x}+\frac{v_0^2}{2}-\frac{1}{x_0}. $$ Then $$ \frac{dx}{dt}=\pm\sqrt{\frac{2}{x}+v_0^2-\frac{2}{x_0}}\,, $$ a first order equation in separated variables whose solution is $$ \int_{x_0}^x\frac{dz}{\sqrt{\frac{2}{z}+v_0^2-\frac{2}{x_0}}}=\pm t. $$ The integral can be computed explicitly, although I do not think you can find a closed form for $x$ in terms of $t$.