https://youtu.be/zXjvuGqrDkQ
PHYS 1101: Lecture Nine, Part Three
And here’s the beginning of our new material for this lecture, the new content. Section 3.3 focuses entirely on what’s called projectile motion. That’s nothing more than any object that’s in free fall. What does free fall mean? I remind you, has to satisfy basically three criteria. It’s an object that has to be in the air, flying through the air. It’s either dropping straight, flying straight up, or it’s been thrown at an angle.
The only real requirement is that when it’s in the air nothing else is touching it. Air resistance is negligible, so the wind isn’t touching it, if you will. It has to be in the air on its own. If it is, no matter what direction it’s headed, it’s a projectile no matter what it is, a rock, a person, a car. When it’s a projectile, here’s what you get. It’s the same type of motion we’ve been describing.
In other words, any projectile, if it’s in two dimensions, if it starts out at some angle and follows a curved arc, it’s nothing more than a special case of what we’ve already looked at, a special case of two dimensional motion.
Here’s what happens. Because of the influence gravity on any object flying through the air, that’s the only thing influencing the motion of the object, the acceleration, the delta-v that occurs every second, is always straight down, has a magnitude of what we call g.
The magnitude g is just a positive number, 9.8 meters per second every second. Just to orient you to how big this is, or what this means, 9.8 meters per second is approximately 20 miles an hour. That means the y component of the velocity, you’re adding about 20 miles an hour to that in this downward direction. You’re stretching that vertical part, the velocity, stretching it down by about 20 miles an hour every second.
With the standard axis definitions, meaning we’re going to go with positive up for y, positive x to the right, a downward constant vector means that the y component for that vector is going to have to be the negative, then the value, or the magnitude, which is 9.8 meters per second squared, and because it’s straight down, Ax has to be 0. Here’s what that means in terms of our two dimensional kinematic problem solving.
The only modifications are that if you have a projectile, you immediately get to write down two things. First, as you’re starting to visualize the problem immediately sketch to remind yourself that your acceleration has to be straight down for this object. When you get to the problem solving step where you’re going to list your knowns, because your acceleration is always this, for free, you write down in your list of knowns that Ax has to be 0, and Ay has to be mines 9.8 meters per second squared.
It has to be minus because of choosing a standard positive in the up direction, and the value has to be g. So all of the problem solving steps we went through last lecture apply. It’s just if it’s a projectile, it immediately becomes somewhat simpler. This has to be the case for a, and these are the implications of that; Ax is 0, and Ay is this.
Here are a couple of quiz questions to just warm you up, and get you appreciating what the implications are of having an acceleration that’s straight down. Let me remind you that this means that our delta-Vy is downward. In other words, we’re stretching the y component all the time, but the x component of the velocity can’t change.
The change in that horizontal part of velocity has to be 0 the whole problem long. This is because our delta-v vector looks a lot like this acceleration vector. It has to be in the same direction. So with that in mind, and thinking about the implications then on how long it takes for an object to fall, I have this classic question for you. In fact, we saw a demonstration of this a few lectures ago.
If we had a cannon here firing a ball from the top of a cliff, it’s firing it straight out horizontally, and at the same instant you drop a blue cannon ball, which of those balls hits the ground first?
The next two quiz questions are another standard problem that people need to think about carefully in the context of a projectile for which acceleration is straight down.
Consider you’ve got a plane flying along at a constant speed. Of course, if it’s flying along that means it’s headed horizontally, and it drops a red package out of its cargo door. Possible trajectories for that package are shown as these purple lines. So immediately the package is dropped, and as time goes on, does it follow this path as it hits the ground, this path, or this path? Before we answer that, I first asked you in Question 6 — where is the plane when the package hits the ground?
So the plane continues on at a constant velocity. Of course the red now is the package, and it’s at the ground level, as opposed to up here where the package was still attached to the plane.
Question 7 then is the standard way that this question is usually asked. If this plane is flying along at this constant velocity, constant speed, and flying along horizontally, what’s the trajectory that the package is going to take, as seen by somebody that’s standing on the ground? Meaning, somebody that’s standing here on the sidewalk watching the plane fly by. What path do they see that package take?
To give you one thing to think about, ask yourself when the package is still in the plane, what is its velocity? And if the package is simply dropped, its velocity initially has to be the same as it was just before it was dropped. Dropping means that you’re not changing the velocity of that instant.
As it’s dropped, the second it is let go, it starts experiencing this acceleration due to gravity, which is straight down. So whatever velocity the package had when it was let go, that velocity starts to change by this little delta-v vector.