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Welcome back.

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I now want to introduce you
really just to a different

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notation for writing vectors,
and then we'll do that same

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problem or a slight variation
on that problem

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using the new notation.

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This is just to expose you to
things, so that you don't get

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confused if your teacher uses a
different notation than what

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I've been doing.

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So when we did the unit vectors,
we learned that we

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can express a vector as a
component of its x- and

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y-components.

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So let's say I had a vector--
let me just pick a random

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vector just to show you.

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So say I had vector a and that
equals 2 times the unit vector

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i plus 3 times the
unit vector j.

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That's the unit vector notation,
and I actually

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looked it up on Wikipedia, and
they actually called it the

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engineering notation.

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That's probably why I used it
because I am an engineer, or I

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was an engineer before
managing money.

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But another way to write this,
and I call this the bracket

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notation, or the ordered pair
notation, is you could also

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write it like this.

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We have this one bracket.

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That's the x-component, that's
the y-component.

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It almost looks like a
coordinate pair, but since

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they have the brackets, you
know it's a vector.

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But you would draw it
the exact same way.

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So given that, let's do
that same problem

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that we had just done.

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Hopefully, this make
sense to you.

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It's just a different
way of writing it.

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Instead of an i and a j, you
just write these brackets.

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Instead of a plus, you
write a comma.

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Let me clear this.

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I'm going to do a slight
variation.

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This was actually the second
part of that problem.

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My cousin gave these
problems to me.

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They're pretty good, so I figure
I'd stick with them.

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So in the old problem, let me
draw my coordinate axes again.

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That's the y-axis.

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That's the x-axis.

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So in the old problem, I started
off with a ball that

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was 4 feet off the ground.

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So let's say that's 4.

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And I hit it at 120 feet per
second at a 30-degree angle.

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So that's a 30-degree
angle like that.

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Its' a 30-degree angle
to the horizontal.

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And there's a fence 350 feet
away that's 30 feet high.

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It's roughly around there.

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That's 30.

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And what we need to do is figure
out whether the ball

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can clear the fence.

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We figured out the last time
when we used the unit vector

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notation that it doesn't
clear the fence.

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But in this problem, or the
second part of this problem,

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they said that there's
a 5 meter per second

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wind gust to the right.

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So there's a wind gust of 5
meters per second right when I

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hit the ball.

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And you could go into the
complications of how much does

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that accelerate the ball?

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Or what's the air resistance
of the ball?

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I think for the simplicity of
the problem, they're just

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saying that the x-component of
the ball's velocity right

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after you hit it increases
by 5 meters per second.

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I think that's their point.

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So let's go back and do the
problem the exact same way

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that we did it the last
time, but we'll

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use a different notation.

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So we can write that equation
that I had written before,

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that the position at any given
time as a function of t is

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equal to the initial position--
that's an i right

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there-- plus the initial
velocity.

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These are all vectors.

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Initial velocity times t plus
the acceleration vector over

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2t squared.

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So what's the initial
position?

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And now we're going to use
some of our new notation.

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The initial position when I hit
the ball, its x-component

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is 0, right?

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It's almost like its coordinate,
and they're not

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that different of a notation.

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And then the y-position is 4.

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Easy enough.

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What's its initial velocity?

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Let me do it.

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So we can split it up into the
x- and the y-components.

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The y-component is 120 sine of
30 degrees and then the x

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component is 120 cosine
of 30 degrees.

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That's just the x-component
after I hit it.

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But then they say there's
this wind gust so it's

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going to be plus 5.

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I think that's their point when
they say that there's

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this wind gust. They say that
right when you hit it, for

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some reason in the x-direction,
it accelerates a

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little bit by 5 meters
per second.

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So the velocity vector.

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This notation actually is
better, because it takes less

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space up, and you don't have
all these i's and j's and

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pluses confusing everything.

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So the initial velocity
vector, what's its

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x-component?

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It's 120 cosine of 30.

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Cosine of 30 is square root of
3/2 times 120 is 60 square

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roots of 3, and then
you add 5 to it.

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So what is that?

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Let me just solve
it right now.

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So 3 times the square root
of 3 times 60 plus 5.

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So let's just round up
and make it easier.

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It's 109 meters per second.

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108.9, so let's just say 109.

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So the x-component of
the velocity is 109.

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And the y-component was just
120 times the sine of 30.

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Well, sine of 30 is 1/2,
so this is 60.

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Oh, sorry, this should be
brackets, although some people

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actually write the parentheses
there so it looks just like

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coordinates, but I like to keep
it with these brackets so

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that you don't think that these
are coordinates since

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you know these are vectors.

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And a position vector is really
the same thing as a

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position coordinate.

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But a velocity vector is
obviously not a coordinate.

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What's the acceleration
vector?

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Well, the acceleration vector,
as we said, goes straight--

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that's not straight down.

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This is straight down at minus
32 feet per second squared.

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That's the acceleration
of gravity on Earth.

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So the acceleration vector
is equal to -- it has no

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x-component and its y-component
is minus 32.

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So now let's put these back
in that original equation.

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So our position vector, and
I'll switch colors to keep

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things from getting
monotonous.

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Our position vector-- these are
little arrows or one-sided

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arrows-- equals my initial
position, and that's 0, 4 plus

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my initial velocity vector,
109, 60 times t, and I'm

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running out of space, plus at
squared over two, so t squared

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over 2 times my acceleration
vector, 0 minus 32.

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This is actually a little
cleaner way of writing it, but

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this is exactly what
we did when we did

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it with unit vectors.

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Instead of writing i's and j's,
we're just writing the

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numbers in brackets here.

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So let's see if we can
simplify this.

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So let me write it in a
different color, so that you

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know I'm doing.

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OK, so our position vector t
is equal to 0, 4 plus-- and

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now we can distribute this t,
multiply it times both of

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these-- plus 109t, 60t plus--
and we can distribute this t

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squared over 2.

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Well, that times 0 is 0.

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And then that times minus
32 is minus 16t squared.

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Now we can add the vectors.

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So the position at any t.

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So let's add all the
x-components of the vectors.

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0, 109t, 0, so we
just get 109t.

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And then what's the
y-components?

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4 plus 60t minus 16t squared.

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And there we go.

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We've defined the position
vector at a

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function of any time.

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So let's solve the problem.

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Now that they have this wind
gust and our x velocity's

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going a little faster, let's see
if we can clear the fence.

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So how long does it take to
get to 350 feet in the

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x-direction?

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Well, this number right
here has to equal 350.

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So we have 109t has to
be equal to 350.

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And so what's 350
divided by 109?

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350 divided by 109 is equal
to 3.2 seconds.

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t is equal to 3.2 seconds.

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And so what's the height
at 3.2 seconds?

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So let's square that.

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3.2 times 3.2 equals times
16 equals 164.

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So this equals 164.

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And then what's 60 times 3.2?

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60 times 3.2 is equal to 192.

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So what do we get?

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We get 192 plus 4 minus
164 is equal to 32.

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So our position vector at time
3.2 seconds is equal to 350

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feet in the x-direction and 32
feet in the y-direction, and

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that will clear that
30-foot fence.

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Our ball's going to be two
feet above the fence.

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Hope I didn't confuse
you too much.

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See you soon.

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