The movement of a ball is a matter of physics, not mathematics. The crucial point is that the downward acceleration, as long as the ball stays "near" the surface of the earth (not tens of miles above) it, is that the acceleration due to gravity remains constant while, ignoring air resistance, there is no acceleration horizontally: the vertical acceleration is -g which is approximately -9.8 m per second per second so the vertical speed, t seconds after the kick, is -9.8t plus the initial speed, so
= -gt+ v_{y0})
, while the horizontal velocity remains the initial horizontal speed:
= v_{x0})
. That means that, in t seconds, the ball goes, horizontally, distance
. Vertically, with changing speed, it's a little harder but with
constant acceleration, we can use the
average speed. With starting speed, vertically,
, and speed after t seconds,
, the average speed will be
t)
and so the ball have gone, vertically, the distance
t\right)t= -(g/2)t^2+ v_{y0}t)
.
That is,
while
t^2+ v_{y0}t)
. To get that in terms of x and y only, from
we get
and then
(x/v_{x0})^2+ v_{y0}(x/v_0)= -\frac{g}{2v_{x0}^2}x^2+ \frac{v_{y0}}{v_{x0}}t)
.
That's what the "coefficient for the squared x" is- it depends upon the acceleration due to gravity and the initial horizontal speed. Yes the graph of "a squared number" goes in an arc- more correctly, a parabola. But I would not say that a cubed number goes "all kinds of crazy"- it just has
two arcs. You probably know that a quadratic equation,
may have
two solutions so its graph has to cross y= 0 twice. Similarly, an equation involving
may have three roots and so its graph may cross y= 0
three times. If it goes up at, say x= 0, it must reach some maximum value and then come back down to cross the axis again, the turn and cross a third time- two arcs instead of one.
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Originally Posted by HallsofIvy 
