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I'm trying to figure out how to model projectile motion which can be defined using a simple [quadratic] equation which defines projectile parabola:

$$y = y_0 + v_{0y}t - \dfrac{1}{2} gt^2$$

I came across some slides from Brown University which model the motion as follows:

$$\begin{align} \text{Matrix form: }& \begin{pmatrix} 0 & 100 & -10\end{pmatrix} \\ \\ \text{state: }& \begin{matrix}y & y' & y'' \\ p & v & a\end{matrix} \\ \\ &\begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 2 & 1 \\ \end{pmatrix} \end{align}$$

But I'm not quite sure how they came up with that transition matrix. Can anyone shed some light how to pick the state variables and how to model the motion?

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He started from $$ y = y_0 + v_0 t + \frac 1 2 a t^2$$ but then for some reason best known to himself decided to take $\frac 1 2 a$ as $y''$ rather than a, (and ignored the fact that his $y''(t)$ is no longer equal to $d y'(t) / dt$, but hey, Lewis Carroll was also a mathematician, and this guy, like Humpy Dumpty, can make words mean whatever he feels like they ought to mean).

So he gets the approximate equations $$\begin{matrix} y(t) \\ y'(t) \\ y''(t) \end{matrix} \approx \begin{matrix} y(t-1) + y'(t-1) \\ y'(t-1) + 2y''(y-1) \\ y''(t-1) \end{matrix}$$ or $$\begin{bmatrix} y(t) & y'(t) & y''(t) \end{bmatrix} \approx \begin{bmatrix} y(t-1) & y'(t-1) &y''(t-1) \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 2 & 1\\ \end{bmatrix} $$ He can't seem to make up his mind whether $y''$ is really a variable, or just the arbitrary constant $-10$, (or was that $+10$, or $\pm 20$ - I've lost the will to care...

It would be more obvious what's going on if you replace the 2's with 1's in his equations, and let $y''(t)$ actually BE the second derivative of $y(t)$, not something close.

Do people actually keep their jobs as university lecturers for teaching like this nowadays? It's 50 years since I was an undergrad, so I don't know....

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  • $\begingroup$ Uh, super interesting. I thought I was missing something fundamental there, but your answer makes me think I probably havent - it just seems kinda all over the place in those slides. I dont think I would ever have thought of that. $\endgroup$ – milosgajdos Apr 2 '19 at 23:46

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