Skip to content

A fun kinematics discovery

June 2, 2011

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:


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

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


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?

2 Comments leave one →
  1. June 4, 2011 11:49 am

    “How doI capture/create this spirit with more of what we do everyday?”

    This question was always on my mind when I was teaching/coaching/managing college entrepreneurs but after reading this question I have not been able to think of anything else. Thank you for making me a little bit crazier.

    I came across a column at where a Dr. Michio Kaku answers physics questions every Wednesday. What if you took fifteen minutes a week to discuss something like this: ? As a physics student, I would love to do this however I’m sure that there would be people who couldn’t care less. I’m just throwing an idea out there, do you know if something like this has produced results?

    • June 4, 2011 10:34 pm

      Thanks for the great ideas and links to Kaku’s column, I had not seen that before. I do try to do something like that where I post interesting links on our class blog and discuss them at the beginning of class, but often, this devolves down into—can you tell us about this or that, and I think was was particularly great about this instance, was the student had a question, and the means and desire to find the answer, rather than simply asking me for an explanation. This is what I want to develop further in my own teaching.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: