Consider an object dropped from a certain position, and the only force is acceleration due to gravity. The object accelerates the same throughout the free fall; not speeding up or slowing down. So this is constant motion, right? Why, then, does the total distance (position) covered show a parabolic shape?
The total distance covered (final position, final height) is:
$$S_f=S_0-\frac{1}{2}at^2.$$
This is a parabolic shape. Why does this constancy take on a parabolic shape?
Is it because the initial velocity is horizontal? Is the initial velocity horizontal?
- For example: An object is dropped from the top of a cliff $630$ meters high. It's height above ground $t$ seconds after it is dropped is $630-4.9t^2.$
The object falls vertically, and so you would consider acceleration due to gravity, which is constant.
I'm confused on the parabolic shape of this position function. Does it have anything to do with $t=0?$
I guess this is a pretty basic question for MSE but I would really like to un-confuse myself.
Thank you!
Do you know calculus? If so, this is very easily answerable. Acceleration is the second derivative of position, so if acceleration is constant, we have $x''(t) = a$. Thus by integrating twice, we have $x(t) = x_0+v_0t+\frac{1}{2}at^2$.