How does a quadratic function model a free-fall ride?

Answer 1

In a free fall ride, the object (car) is projected, at time 0, with some initial velocity. Suppose that velocity has a horizontal component of h metres per second and a vertical component of v metres per second.

It is assumed that the only force acting on this body is that of gravity. Air resistance is assumed to be insignificant. Consider the position of the car at time t seconds after "launch", measured from the launch position.

There are no forces acting on it in the horizontal direction. So, following Newton's law, it continues to travel in the horizontal direction at h metres per second. And therefore, its position is ht metres from the launch position.

In the vertical direction, the force is a downward acceleration of g metres per second2. So the vertical position at time t, is vt - 1/2*gt2.

Thus x = ht

and y = vt - 1/2*gt2

From the first equation, t = x/h. Substituting this in the second gives

y = v*x/h - 1/2*g(x/h)2 = -(1/2*g/h2)x2 + (v/h)*x

that is, y = Ax2 + Bx for some constants A and B

[specifically A = -(1/2*g/h2), and B = (v/h), but the exact values are not relevant here.]

Now y = Ax2 + Bx is the quadratic function model, therefore it is a suitable model for such motion.