Hey everyone, I’m Felipe. Today, I’m going to show you how to use Desmos to make a physics problem easier to understand.

I think learning tools should be easy for students to access, especially when we’re trying to figure something out on our own.

I sometimes find it hard to picture what a physics equation actually means. Seeing a graph helps me connect the numbers to what’s happening.

I’m studying Computer Science and Mathematics at Cégep de Sherbrooke. We take courses like programming, calculus, and physics. This program prepares us for university studies in fields like computer science and engineering.

My goal is to show you how to find a ball’s maximum height and the time it takes to hit the ground.

You can open Desmos in another tab and follow along with me.

We’ll also see why an answer can be mathematically correct but still not make sense for our problem.

In physics assignments, we often use equations to describe movement.

But if we don’t understand what our answers represent, we can choose the wrong one.

Imagine throwing a ball straight up. We want to know how high it goes and when it lands.

I’ll show you how to use a graph to find those answers and check what they mean.

For this tutorial, I’m using Desmos, a website suggested to me while I was preparing this presentation.

On the left-hand side of the screen, you’ll see the area where you enter equations. The graph appears on the right-hand side.

[Point to each section with your mouse.]

The plus and minus icons let you zoom in and out. You can also use your mouse to drag the graph around.

If you want a more precise view, open the graph settings menu using the wrench icon.

[Open the graph settings.]

For our example, I’ll set the horizontal axis from negative one to three, and the vertical axis from negative two to ten.

[Enter the bounds, then close the settings.]

That gives us enough room to see the important parts of our graph.

Let’s say the ball starts one point five metres above the ground, moving upward at twelve metres per second. We’ll ignore air resistance.

First, left-click the first empty expression line and enter this equation.

[Type: y=-4.9x^2+12x+1.5]

That’s y equals negative four point nine x squared, plus twelve x, plus one point five.

Here, x represents time in seconds, and y represents height in metres. The negative term comes from gravity.

The curve goes up, reaches a peak, and comes back down. This shows how the height changes over time. It isn’t a picture of the ball’s path through the air.

[Select the curve, then its highest point.]

To find the maximum height, click the highest point on the curve.

The coordinates are approximately one point two two and eight point eight five. That means the ball reaches a height of eight point eight five metres after about one point two two seconds.

Now, when does it hit the ground?

The ground is where the height equals zero. So we need to find where the curve crosses the horizontal axis.

[Select the crossing to the right of zero.]

This point tells us that the ball lands after about two point five seven seconds.

[Point to the crossing to the left of zero.]

There’s also a crossing at a negative time. It belongs to the equation, but it happens before our starting time. We can’t use it as the landing time after the throw.

We wanted to find the ball’s maximum height and when it lands. Using Desmos, we found about eight point eight five metres and two point five seven seconds. The graph also helped us understand which answer to use.

Try changing the starting velocity from twelve to fifteen and see what happens to the graph. Thanks for watching, and subscribe for more tutorials like this!

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Total MP3 length: 00:00

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Error density: 2%