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.
- Why is $a$ halved? It goes from $9.8$ to $4.9$.
- Why is the time $t$ squared?
These are basic equations, however, I couldn't find explanations as to the whys, only methodology telling me to "plug in".
Thank you.
Best Answer
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.