Physics and g = 2s t^-2

[Silverthorn]

edit: eh, looks like I was too slow again

You can also solve it via a very simple differential equation

a is the derivative of the velocity v, therefore: v = Integral(a dt) = a*t + C[sub]1[sub]
We'll just omit C[sub]1[sub] because we assume that the object doesn't have any initial movement.
The velocity v is the derivative of the position s, therefore: s = Integral(v dt) = 1/2*a*t² + C[sub]2[sub]
Let's say we start from the origin, so that this C[sub]2[sub] becomes zero as well.
Plug in Earth's acceleration g for a and we'll get: s = 1/2*g*t²
When you now bring g to the left side and s to the right, you get: g = 2s*t^-2

We all know gravity is acceleration, and acceleration is only "s t^-2"

That is correct for units. Acceleration is distance per times squared. You won't get any information about potential factors that might pop up randomly (compare to Coulomb's law where a sudden 4*π falls from the sky; same for Schrödinger's equation with ħ/(2m)). Most of the time, it works, but you have to know where to pay attention to special cases