Projectile motion with ordered set notation

Welcome back. I now want to introduce you really just to a different notation for writing vectors, and then we'll do that same problem or a slight variation on that problem using the new notation. This is just to expose you to things, so that you don't get confused if your teacher uses a different notation than what I've been doing. So when we did the unit vectors, we learned that we can express a vector as a component of its x- and y-components. So let's say I had a vector-- let me just pick a random vector just to show you. So say I had vector a and that equals 2 times the unit vector i plus 3 times the unit vector j. That's the unit vector notation, and I actually looked it up on Wikipedia, and they actually called it the engineering notation. That's probably why I used it because I am an engineer, or I was an engineer before managing money. But another way to write this, and I call this the bracket notation, or the ordered pair notation, is you could also write it like this. We have this one bracket. That's the x-component, that's the y-component. It almost looks like a coordinate pair, but since they have the brackets, you know it's a vector. But you would draw it the exact same way. So given that, let's do that same problem that we had just done. Hopefully, this make sense to you. It's just a different way of writing it. Instead of an i and a j, you just write these brackets. Instead of a plus, you write a comma. Let me clear this. I'm going to do a slight variation. This was actually the second part of that problem. My cousin gave these problems to me. They're pretty good, so I figure I'd stick with them. So in the old problem, let me draw my coordinate axes again. That's the y-axis. That's the x-axis. So in the old problem, I started off with a ball that was 4 feet off the ground. So let's say that's 4. And I hit it at 120 feet per second at a 30-degree angle. So that's a 30-degree angle like that. Its' a 30-degree angle to the horizontal. And there's a fence 350 feet away that's 30 feet high. It's roughly around there. That's 30. And what we need to do is figure out whether the ball can clear the fence. We figured out the last time when we used the unit vector notation that it doesn't clear the fence. But in this problem, or the second part of this problem, they said that there's a 5 meter per second wind gust to the right. So there's a wind gust of 5 meters per second right when I hit the ball. And you could go into the complications of how much does that accelerate the ball? Or what's the air resistance of the ball? I think for the simplicity of the problem, they're just saying that the x-component of the ball's velocity right after you hit it increases by 5 meters per second. I think that's their point. So let's go back and do the problem the exact same way that we did it the last time, but we'll use a different notation. So we can write that equation that I had written before, that the position at any given time as a function of t is equal to the initial position-- that's an i right there-- plus the initial velocity. These are all vectors. Initial velocity times t plus the acceleration vector over 2t squared. So what's the initial position? And now we're going to use some of our new notation. The initial position when I hit the ball, its x-component is 0, right? It's almost like its coordinate, and they're not that different of a notation. And then the y-position is 4. Easy enough. What's its initial velocity? Let me do it. So we can split it up into the x- and the y-components. The y-component is 120 sine of 30 degrees and then the x component is 120 cosine of 30 degrees. That's just the x-component after I hit it. But then they say there's this wind gust so it's going to be plus 5. I think that's their point when they say that there's this wind gust. They say that right when you hit it, for some reason in the x-direction, it accelerates a little bit by 5 meters per second. So the velocity vector. This notation actually is better, because it takes less space up, and you don't have all these i's and j's and pluses confusing everything. So the initial velocity vector, what's its x-component? It's 120 cosine of 30. Cosine of 30 is square root of 3/2 times 120 is 60 square roots of 3, and then you add 5 to it. So what is that? Let me just solve it right now. So 3 times the square root of 3 times 60 plus 5. So let's just round up and make it easier. It's 109 meters per second. 108.9, so let's just say 109. So the x-component of the velocity is 109. And the y-component was just 120 times the sine of 30. Well, sine of 30 is 1/2, so this is 60. Oh, sorry, this should be brackets, although some people actually write the parentheses there so it looks just like coordinates, but I like to keep it with these brackets so that you don't think that these are coordinates since you know these are vectors. And a position vector is really the same thing as a position coordinate. But a velocity vector is obviously not a coordinate. What's the acceleration vector? Well, the acceleration vector, as we said, goes straight-- that's not straight down. This is straight down at minus 32 feet per second squared. That's the acceleration of gravity on Earth. So the acceleration vector is equal to -- it has no x-component and its y-component is minus 32. So now let's put these back in that original equation. So our position vector, and I'll switch colors to keep things from getting monotonous. Our position vector-- these are little arrows or one-sided arrows-- equals my initial position, and that's 0, 4 plus my initial velocity vector, 109, 60 times t, and I'm running out of space, plus at squared over two, so t squared over 2 times my acceleration vector, 0 minus 32. This is actually a little cleaner way of writing it, but this is exactly what we did when we did it with unit vectors. Instead of writing i's and j's, we're just writing the numbers in brackets here. So let's see if we can simplify this. So let me write it in a different color, so that you know I'm doing. OK, so our position vector t is equal to 0, 4 plus-- and now we can distribute this t, multiply it times both of these-- plus 109t, 60t plus-- and we can distribute this t squared over 2. Well, that times 0 is 0. And then that times minus 32 is minus 16t squared. Now we can add the vectors. So the position at any t. So let's add all the x-components of the vectors. 0, 109t, 0, so we just get 109t. And then what's the y-components? 4 plus 60t minus 16t squared. And there we go. We've defined the position vector at a function of any time. So let's solve the problem. Now that they have this wind gust and our x velocity's going a little faster, let's see if we can clear the fence. So how long does it take to get to 350 feet in the x-direction? Well, this number right here has to equal 350. So we have 109t has to be equal to 350. And so what's 350 divided by 109? 350 divided by 109 is equal to 3.2 seconds. t is equal to 3.2 seconds. And so what's the height at 3.2 seconds? So let's square that. 3.2 times 3.2 equals times 16 equals 164. So this equals 164. And then what's 60 times 3.2? 60 times 3.2 is equal to 192. So what do we get? We get 192 plus 4 minus 164 is equal to 32. So our position vector at time 3.2 seconds is equal to 350 feet in the x-direction and 32 feet in the y-direction, and that will clear that 30-foot fence. Our ball's going to be two feet above the fence. Hope I didn't confuse you too much. See you soon.