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## Two-dimensional projectile motion

Current time:0:00Total duration:10:17

# Projectile motion with ordered set notation

## Video transcript

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.