https://youtu.be/IL-Dx8mxyW0
PHYS 1101: Lecture Seven, Part Eight
Let me show you a movie now where you can visually appreciate even more this disconnect between the vertical and horizontal motion. This movie shows an apparatus where on one hand it can let go and just drop the ball directly. On the other side, a ball can be mounted that will be shot out at the same time. Both of these objects will become projectiles when they free fall as soon as they let go. The acceleration that they experience is straight down. It’s acceleration due to gravity.
When you let these two objects go, they follow very different paths, but what they’re going to show you as they slow down is that these objects are falling at the same rate. If you are to look at the motion diagram you’d see that the ball that falls straight down is in free fall. It’s just like dropping a rock. Its velocity gets bigger and bigger by nine point eight meters per second every second.
The orange or yellow ball, when it’s fired, it starts out with a velocity that’s horizontal. Its vertical velocity starts out zero just like this one does. It’s dropped. It starts from rest. It’s only the vertical part of the velocity that gets bigger and bigger by nine point eight meters per second every second.
The horizontal motion, if you were to look at the shadow that it casts horizontally, you would see that it is plotting along at the same velocity that it was kicked off with here. The horizontal part to the velocity is what causes the ball to travel to the right this distance during that time. The time it takes to fall is vertical motion and that’s only impacted by the acceleration due to gravity which is straight down. These two objects are going to hit the ground at the same time. And there, they do.
Here’s a quick sketch of that same movie that you saw with “x” and “y” axes. And question fourteen to you is, the time that it takes the projectile to hit the ground is set by what? The “Y” part of the motion only, the “x” part of the motion only, or both?
Let me add to this sketch that for any projectile the acceleration vector is straight down. For a projectile, this vector that’s straight down, it only has a “y” component. “Ay” is minus nine point eight meters per second squared. “Ax” is zero. This is what keeps the “x” part of the velocity the same.
This next quiz question perhaps is easy for you to answer then having discussed all that. After what I have told you about “Ay” as we looked at the shadow in this direction and thinking about the shadow in this direction, to look at the velocity “Ax” information and “Ay” information. What’s the direction of “A” for this purple object that underwent this trajectory?
Here’s a picture to summarize it again for you. We’re going to be studying motion like this that is trajectory. We’re going to have an initial velocity vector, a final velocity vector, and acceleration. I’m going to be explicit here. This is an example of what’s called a projectile. I’ll define that more clearly for you in a minute.
The real life quantities are these bold vectors that I have drawn for you. There’s going to be an initial speed and direction to define this vector. There’s going to be acceleration at some magnitude and direction. In our case, this one we’re drawing it straight down. And then there is going to be some final speed and direction.
In real life, these vectors are what your eye is drawn to. It’s the hypotenuse information. But in order to solve the math and to do the problem, we have to break these hypotenuses, I’m not sure if that’s the right word, break these vectors up into their components. We have to work with the horizontal parts. The vertical parts, got to strip out those parts from the vectors.
The variables “t” naught and “t,” they’re common to both because the motion is happening at the same time, horizontally and vertically. Here again is a summary of all of the mathematical variables we’re going to work with to do this two dimensional analysis.
Question 16 for you, at the highest point in this trajectory, what are the components of “v” given positive “y” up, positive “x” to the right? Just a summary statement here that the trajectory we’ve been watching is called a projectile. We have a lot of experience with that day to day.
Clearly, what defines a projectile? It really is the same as an object in free fall. The criteria are that it has to be in the air. It’s moving through the air, nothing is touching it. Everything we’re going to consider in this class we can treat as ignoring air resistance, that the wind, the effect of the air bombarding it as it moves along, it can be ignored. Air resistance negligible. Nothing is touching it. It’s in the air on the planet Earth. It’s a projectile. And that brings me to the end of Lecture 7.