$$(\dot{r})^2 = \frac{2 \mu}{r} + 2h$$
Where mu and h are constants.
I have no idea how to solve it, maybe there is a trick I didn't know.
The only thing that came in mind is to integrate $$\int \frac{dr}{\sqrt{\frac{2 \mu}{r} + 2h}} = \int dt$$ but I don't think this is really a solution, since I don't know too how to evaluate the integral in terms of elementary functions.
So, any tips?
Rewrite the integral as
$$\frac{1}{\sqrt{2}}\int \sqrt{\frac{r}{\mu+hr}}\:dr$$
and use the substitution $r = \frac{\mu}{h}\sinh^2\rho$
$$\sqrt{\frac{\mu^2}{2h^3}}\int2\sinh^2\rho\:d\rho = \sqrt{\frac{\mu^2}{2h^3}}\int\cosh2\rho-1\:d\rho = \sqrt{\frac{\mu^2}{2h^3}}\left(\sinh\rho\cosh\rho-\rho\right)$$
Then undo the substitution with
$$\sinh \rho = \sqrt{\frac{hr}{\mu}} \implies \cosh\rho = \sqrt{1+\frac{hr}{\mu}}$$
giving an equation
$$\sqrt{\frac{hr}{\mu}}\sqrt{1+\frac{hr}{\mu}}-\sinh^{-1}\left(\sqrt{\frac{hr}{\mu}}\right) = \frac{\sqrt{2h^3}}{\mu}(t+C)$$
This is of course only the case where an object keeps moving further and further away from a gravitational body because of your choice for $\dot{r}$ to be strictly positive.