My eighth-grade physics teacher taught us that the equation for height ($h$) during a free fall in a vacuum is equal to the initial height ($H_0$) minus a constant ($G$, though not the $9.8$ meters per second we know and love) multiplied by time ($t$) squared, divided by $2$.
However, $dh\over dt$ is velocity ($v$), and $dv\over dt$ is acceleration ($a$). If acceleration were constant, then his formula would be fine. But acceleration is not constant. It's $G\over h^2$. So if acceleration changes, then the velocity is not a perfect quadratic equation, meaning the equation he taught us is wrong. Presumably he was dumbing it down for $13$-year-olds.
What is the actual equation? How could you arrive at it? I tried setting up a second-order differential equation, but got stuck on $$dh = G\int {1\over h^2}dt dt$$ In order to solve this problem I would need to express $h$ in terms of $t$, which is the very problem I am trying to solve. How does one get around this? I am in Calc BC, so I would love for a step-by-step solution.
