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Still learning how to solve projectile motion problems after 20 years of study

February 16, 2012

To me, one of the most enjoyable things about teaching physics is the opportunity to continually learn new things. I’ve now been studying physics for over 20 years (with about half of that time in some formal setting), and so it might be easy to think that, especially when it comes the basic stuff like projectile motion, there literally is nothing left to learn. Not true. And in the rest of this post I want to show you a beautiful, visual way of understanding projectile motion I learned from two excellent physics teachers, Brian Frank and Frank Noschese. Here’s the link to Brian Frank’s post on this topic that inspired this work.

Average velocity—the most important velocity of all

I’ve often told my students that average velocity is the sort of “all-knowing” velocity. If you know your average velocity, you can easily find an answer to he question “how much longer will this trip take?” and many others. And the power of average velocity isn’t limited to long trips along the interstate to grandma’s house, it works for all situations—even the two-dimensional problems where we rarely use this concept.

Here’s an example. Start with a person who walks three different displacements—first to Annie’s house, then to Bill’s, then to Charlie’s, as shown in this vector illustration below.

The red vector represents the total displacement \vec{\Delta{r}}_{tot}, and we know the fundamental definition of average velocity is:


Or rearranging,

\Delta{\vec{r}_{tot}}= \vec{v}_{avg}\Delta{t}

Average velocity is the vector that when scaled by the time of your trip, gets you to your destination.

Let’s see that on a map. Say I want to go from Atlanta to New York. Wolfram Alpha quotes the direct distance as 741 miles. But of course, if you want to drive from Atlanta to New York, you have to take roads, and they don’t follow a straight line path between the cities.

3 different possible routes to NYC. All will have the same displacement, and therefore the same direction for average velocity. But since the route through Toledo is likely to take much longer, the average velocity for that trip should be much shorter.

Finding average velocity for accelerated motion

Here’s a pretty basic kinematics problem:

Suppose an object is moving at 6 m/s and accelerates uniformly to a velocity of 14 m/s over a period of 4 seconds. What is the average velocity for the entire trip?

Students almost instantly will calculate

v_{avg}=\frac{v_i+v_f}{2}=\frac{ 6\;\frac{\textrm{m}}{\textrm{s}}+14\;\frac{\textrm{m}}{\textrm{s}}}{2}=10\;\frac{\textrm{m}}{\textrm{s}}.

But one can gain much more insight with a graph:

The area of the trapezoid and red box are the same. This is the idea of average velocity—it is the velocity that you could travel at for the duration of the trip, and arrive at the same location as the accelerated object.

But what do you do in two-dimensional problems, where you can’t draw a single velocity vs time graph? You can still do graphical vector constructions. The key idea is the definition of acceleration:

\vec{a}_avg=\frac{\vec{\Delta v}}{\Delta t}

rearranging gives us
\vec{\Delta v}=\vec{a}_{avg}\Delta{t}

The change in velocity is just the acceleration vector, scaled by the time. And for constant acceleration, \vec{a}_{avg}=\vec{a}. In essence, the change in velocity vector, \vec{\Delta v} is a sort of clock in the problem that grows longer as time passes. And we can find the final velocity vector by just adding the change in velocity vector.

Let’s start with a simple example. Suppose a projectile is launched at 20 m/s at an angle of 45°. Find the final velocity after it has been in the air for 2.8 seconds.

In two seconds, the change in velocity will be:

\vec{\Delta{v}}=\vec{g}\Delta{t}=10\;\frac{\textrm{m}}{\textrm{s}^2}\cdot2\;\textrm{s}=28\;\frac{\textrm{m}}{\textrm{s}}\textrm{, downward}

and graphically, we can see:

A diagram of the initial velocity, change in velocity and final velocity.

To find the average velocity, we can use the formula that holds true from 1-d motion as a vector equation (assuming constant acceleration:


We just need to add the initial and final velocity vectors, and then half the length of the resultant to find the average velocity.

For constant acceleration, the average velocity is just the sum of the initial and final velocity vectors, scaled by a factor of 1/2.

One interesting result we see from this is that the average velocity vector reaches the midpoint of the change in velocity vector. This should make some sense, since the average velocity must be exactly “between” the initial and final velocity.

Putting this to use to solve projectile motion problems

We should now be able to put this idea to use to solve projectile motion problems. Remembering the key ideas about average velocity:

  1. The average velocity points in the direction of the final position.
  2. The average velocity touches the midpoint of the change in velocity vector.

Problem 1: A projectile is launched at 20m/s at 45° on horizontal ground. Find where it lands.

Here, we know the average velocity vector must be horizontal. For the average velocity to be horizontal, we know that the final velocity must point 45° below the horizontal.

Doing the graphical measurements, we find that \vec{v}_{avg}=14.2\;\frac{\textrm{m}}{\textrm{s}}\textrm{, to the right}

And we know from previous calculations the time required for velocity to change enough to produce a final velocity of 20m/s at 45° below the horizontal is 2.8s.

Graphical solution to the projectile problem 1

So the total distance traveled by the ball is:

\vec{\Delta x}=\vec{v}_{avg}\Delta{t}=14.2\;\frac{\textrm{m}}{\textrm{s}}\cdot 2.82\;\textrm{s}=39.7\;\textrm{m}

But wait, there’s more. This method can solve much more complex problems as well.

Problem 2: Find the maximum height reached by the above projectile.

We know that when the projectile reaches its maximum height, \vec{v}_f will be horizontal, so we can find the time by measuring the length of the change in velocity vector, and dividing it by 9.8\;\frac{\textrm{m}}{\textrm{s}^2}. Then we determine the average velocity, and scale the average velocity by the time. Graphically, here’s what it looks like:

Finding the maximum height reached by the projectile.

From our construction, the average velocity is:
\vec{v}_{avg}=15.7\;\frac{\textrm{m}}{\textrm{s}}\;\textrm{at }26.5^{\circ}\textrm{ above the horizontal}

and scaling this by the time tells us
\vec{\Delta{r}}=\vec{v}_{avg}\Delta{t}=15.7\;\frac{\textrm{m}}{\textrm{s}}\cdot 1.41\;\textrm{s}=22.1\;\textrm{m}\;\textrm{at }26.5^{\circ}\textrm{ above the horizontal}

In components:
\vec{\Delta{r}}=\left\langle 19.7, 9.86\right\rangle \textrm{m}

This is in good agreement with the traditional solution, provided by Wolfram Alpha.

And there’s even more—we can solve the gnarliest of projectile motion problems—those that used to require the dreaded quadratic formula.

Problem 3: A projectile is launched at 20 m/s at a 45° angle from a 30 m cliff. Find the place where the projectile lands.

This problem is a bit more tricky—we don’t know the final velocity, the final location of the projectile, or the time of flight. Still, we can do it.

The basic idea of this approach still holds true:

  1. The average velocity points in the direction of the final position.
  2. The average velocity touches the midpoint of the change in velocity vector.

In this case, we have one other critical piece of information—the projectile lands on the ground, 30 m below its starting point. When we take the average velocity vector and we scale it by the time in the air, it must reach the ground. This constraint limits us to one solution.

It would be hard figure out the solution to this problem by hand—you’d essentially be thinking something like “When I multiply the average velocity vector (which is the vector between the starting point and midpoint of \Delta v) by the time (1/10 the length of \Delta v), I need to get to the ground. But this is a piece of cake for Geogebra. Here’s what it looks like:

Solution to the 3rd problem, generated in Geogebra.

Unfortunately, Wolfram Alpha can’t spit our a quick solution to this problem, so I have to do the algebra:

First find \Delta t by analyzing the y-motion:
\Delta x = v_{0y}t+\frac{1}{2}gt^2

Solving this quadratic gives us t=4.3\;\textrm{s}

Now use the horizontal velocity to find the final position:

\Delta x=v_{0x}t=20\;\frac{\textrm{m}}{\textrm{s}}\cos{45^{\circ}}t\cdot 4.3\;\textrm{s}=60.8\;\textrm{m}


If you want to play with my Geogebra construction to solve other projectile motion problems, here’s a link. Warning—I am by no means an expert in Geogebra, so my file is more of a hack. I’m sure a pro like Dan Cox could turn it into a wonderfully elegant physics homework solver that would slice and dice every projectile motion problem under the sun.

Should you teach this to students?

First, I just want to marvel in how awesome it is to be able to do fairly difficult projectile motion problems without ever needing components or the quadratic formula. I also love, love, love how much this method really builds on the critically important idea of average velocity.

I also value how much intuition this method gives into how the velocity vector changes through its flight, and I feel like it gives a greater sense of meaning to the quantities you’re dealing with—displacements, velocities, and changes in velocities—it reminds you of the vector nature of these objects. I think if I spent a bit more time with this, and paired with some sort of animation of the velocity vector as it moves through the air, and then put this initial/final vector analysis on top of the animation of the projectile moving through the air, it could help students develop a deeper understanding of the underlying physics of projectile motion, without having to constantly worry about which of the nasty kinematics equations to use.

At the same time, vectors are one of the most confusing topics for introductory physics students, and the tendency to drop vector notation and treat everything as a scalar is a strong one, so I worry there are a ton of places in this approach where students could easily go wrong, which makes me pretty hesitant to really try this out with my freshmen.

Still, the coolest thing about working on this for the past couple of days has been to discover an entirely new approach to something I thought I’d mastered years ago and felt I had nothing left to learn about. The biggest question I have is how do we help students (and some faculty) to see that there are moments like this waiting for them in whatever they may choose to study—if they are willing to seek them out.

5 Comments leave one →
  1. February 16, 2012 10:23 am

    “I also love, love, love how much this method really builds on the critically important idea of average velocity. ”

    I do too. I essentially ignored the concept of averge velocity as a problem solving technique for 4 years, until a student showed the class how he used it to easily understand and solve some problems. In my AP Physics class a lot of students feel more comfortable using plug and chug with the other kinematics equations, and I’m not sure how hard to push them to look for average velocity shortcuts. This is a conversation I’d like to have with our Physics 1 teachers now.

  2. February 16, 2012 11:30 am

    I’m glad someone was able to write this up. All I had time to do was to throw up a picture without explaining it.

    The other way to solve the problem with the 30 meter drop is to take it 1 second at a time, updating the location of the particle. The only tricky part is the last step where you need it to land at exactly 30 meters. The first 4 seconds gets you the most of the way, and you have to be a little smart about figuring out how much time is needed to for the last step.


  1. Still learning how to solve projectile motion problems after 20 years of study | PhysicsLearn |
  2. Graphical methods using average velocity « Gas station without pumps
  3. Graphical methods using average velocity « Gas station without pumps

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