During our review a couple of weeks ago, one of my honors physics classes was looking at projectile motion, and the always intimidating equation:

$\Delta y= v_{0y}t+\frac{1}{2}gt^2$

I reminded them that they can see this from velocity vs time graph.

Previously, we found that you can think of this graph as two ways of looking at a PB&J sandwich:

and

But this time, a curious student said, shouldn’t we be able to calculate the areas of the two triangles directly, and see that this equation works?

I said “yes, but I’ve never tried it—let’s do it and see what we find out.”

And this began a wonderful half hour session where our 10 person class split into 3 groups and tried different approaches to the problem on the whiteboard.

As best I remember it, here’s the solution we came up with.

Let’s start with a general projectile that is in the air for $t$ seconds. We won’t limit it to cases where the y-displacement is zero.

We need to start by finding the time when the projectile reaches the peak of its motion $t_{peak}$. We can do this by noting that the y-velocity is zero at that instant. So given that
$v_y=v_{y0}+gt$

We can set $v_y = 0$, and solve for $t_{peak}$

$t_{peak}=-v_{y0}/g$

This makes sense—the units are correct, and the negative in front of the expression will cancel out the negative in g, giving us a positive expression.

Now our general graph looks something like this:

And so we can find the displacement by adding the areas:

$\begin{array}{rcl} \Delta y &=&Area_{red}+Area_{blue}\\ \Delta y&=& \frac{1}{2}\left(-v_{y0}t_{peak}\right)+\frac{1}{2}g\left(t-t_{peak}\right)^2\\ \end{array}$

And substituting in our previous expression for $t_{peak}$ gives us
$\Delta y = \frac{1}{2}v_{y0}\left(-\frac{v_{y0}}{g}\right)+\frac{1}{2}g\left(t+\frac{v_{y0}}{g}\right)^2$

Expanding the square, we get
$\begin{array}{rcl} \Delta y &=& \frac{1}{2}v_{y0}\left(-\frac{v_{y0}}{g}\right)+\frac{1}{2}g\left(t^2+\frac{2v_{y0}t}{g}+\frac{v_{y0}^2}{g^2}\right)\\ &=& -\frac{v_{y0}^2}{2g}+\left(\frac{gt^2}{2}+v_{y0}t+\frac{v_{y0}^2}{2g}\right)\\ \Delta y &=& \frac{1}{2}gt^2+v_{y0}t\\ \end{array}$

Bingo! With a bit of algebra, we showed that you can indeed show that two different approaches to the same graph do lead to identical results. When my students completed this, after much gnashing of teeth with algebra, and stopping one another to re-explain a particular step, all of them came away very impressed by this discovery and with a tremendous feeling of accomplishment, and I’m not completely sure why. I think it must in some part be due to the fact that we were simply “playing” with the ideas—I didn’t know if it was possible, and said is much to the class. Harnessing the power of play and exploring the unknown was a fun way to see what is a pretty boring topic in a rather new light, and highlight the powerful of mathematics at the same time. How doI capture/create this spirit with more of what we do everyday?