The final distance of an object dropped from a certain height is:
$$S_f=S_0-\frac{1}{2}at^2,$$
$S_f=$ Final distance
$S_0=$ Initial height from which the object was dropped
$a=$ acceleration due to gravity (gravitational acceleration)
$t=$ the time traveled by object.
These are basic equations, however, I couldn't find explanations as to the whys, only methodology telling me to "plug in".
Thank you.
On
We consider an object of mass $m$ dropped.
Newton's second law: $F=ma$.
Newton's law of gravitation: $F = G\displaystyle\frac{mM}{r^2}$
Equating the forces gives $a=\displaystyle\frac{GM}{r^2}$
As an aside, this proves that gravitational acceleration is independent of the mass of the object (since $G$ is constant and $M$ is the mass of the earth, also constant), reminiscent of the famous Pisa experiment.
We also note that gravitational acceleration in this context is [approximated by] a constant $-a$, so solving the IVP $$\begin{cases}x''=-a\\ x'(0)=0\\ x(0)=S_0\end{cases}$$ gives $x(t)=S_0-\displaystyle\frac a2 t^2$. The halving is just a mathematical consequence of the differential equations, so is the time's square.
On
About the halving:
When the speed is constant, $s=vt$ is obvious.
When you accelerate or decelerate between $0$ and $v$, it is also obvious that $s<vt$. And more precisely, if the growth or decay is linear, you get exactly the half (intuitively because you are as much "slow" as you are "fast").
About the squaring:
The speed is $\propto t$, hence by the above reasoning, $s\propto vt\propto t^2$.
Acceleration is the second time derivative of position. If it's constant, we have: $$\frac{d^2}{dt^2} x = a_0$$ Integrate both sides: $$\frac{d}{dt} x = a_0 t + v_0$$ If we integrate both sides again, we get: $$x = \frac{1}{2} a_0 t^2 + v_0 t + x_0$$
The $v_0$ and $x_0$ come from the constants of integration, and are given by the initial conditions.
Another reason there's a one half: after time $t$ has passed, the velocity is $a_0 t + v_0$. But it hasn't been going that speed the entire time, it started at $v_0$. So the position must be less than $v(t) t + x_0 = a_0 t^2 + v_0 t + x_0$. That the coefficient is exactly one half should be shown more rigorously, but at least we know it's less than one.