How might I go about writing a parabolic equation in standard form $ax^2 + bx + c$ given all of the following measurements:

$X_0, X_f, Y_o, Y_f$: the initial and final x and y positions.
$V_{x0}, V_{xf}, V_{y0}, V_{yf}$: the initial and final x and y velocities.
$t_x, t_y$: the time it takes for $x_0$ to transition to $x_f$; same for y. These are equal in parabolic equations.
$a_x, a_y$: acceleration in x and y.

I don't know how I would give the trajectory as a normal parabola from this data because I'm not sure which variables to substitute and where into $ax^2 + bx + c$.

  • $\begingroup$ What do $t_x, t_y$ represent? Typically you would express both x and y as functions of t, which can be massaged into an implicit relation between x and y. $\endgroup$ – Asad Saeeduddin Nov 22 '13 at 2:40
  • $\begingroup$ What do any of the variables you cited represent? There are 12 of them, and you didn't explain what any of them are, which naturally makes things rather difficult for people who want to answer. $\endgroup$ – DumpsterDoofus Nov 22 '13 at 2:41
  • $\begingroup$ Information has been added to explain the variables. $\endgroup$ – person27 Nov 22 '13 at 4:29

You may be overthinking it. In general when you need to solve for coefficients you first need to ask what relates the coefficients? In this case it doesn't seem you've written precisely what equation you want. In your post, you've written the parabola in terms of x. Do you mean you want y position as a function of x position? This isn't the standard way to go about it, normally the most useful method is to give x and y in terms of t.

If the parabolic equation gives the height of the ball at time t: $y(t)=a t^2+b t+c$ and a linear equation gives the position of the ball at time t: $x(t)=d t$, and if we assume gravity is directed downwards giving an acceleration of $g$ we can clearly see that $d=V_{x0}=V_{xf}=(X_f-X_0)/(t)$. (gravity does not change the horizontal velocity at all)

Now, gravity dictates the ball's acceleration completely, which tells us $a=-\frac{g}{2}$. That leaves two constants, and so we need to come up with two equations to solve for them. If we are given the initial and final positions and the time taken to go between the states, we don't need velocities at all. The two equations are:

$$Y_0=y(t_0)=a t_0^2+b t_0+c$$ $$Y_f=y(t_f)=a t_f^2+b t_f+c$$

or, written more clearly for our purposes:

$$b t_0+c=Y_0-a t_0^2$$ $$b t_f+c=Y_f-a t_f^2$$

(written in this form because it's a standard format in linear algebra for a system of linear equations and helps isolate the unknowns)

These simultaneous equations can be solved for $b$ and $c$ in terms of the other components. (try subtracting one equation for the other), giving us all the coefficients.

If you wanted to solve for $y$ in terms of $x$, first solve for $t$ in terms of $x$ and then substitute that value of $t$ into the equation for $y$.


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