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Welcome back.

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I'm not going to do a bunch of
projectile motion problems,

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and this is because I think
you learn more just seeing

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someone do it, and thinking
out loud,

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than all the formulas.

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I have a strange notion that I
might have done more harm than

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good by confusing you with a lot
of what I did in the last

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couple of videos, so hopefully
I can undo any damage if I

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have done any, or even better--
hopefully, you did

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learn from those, and we'll
just add to the learning.

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Let's start with a
general problem.

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Let's say that I'm at the top
of a cliff, and I jump--

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instead of throwing something,
I just jump off the cliff.

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We won't worry about my motion
from side to side, but just

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assume that I go
straight down.

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We could even think that someone
just dropped me off of

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the top of the cliff.

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I know these are getting kind
of morbid, but let's just

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assume that nothing
bad happens to me.

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Let's say that at the top
of the cliff, my initial

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velocity-- velocity initial-- is
going to be 0, because I'm

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stationary before the person
drops me or before I jump.

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At the bottom of the cliff
my velocity is

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100 meters per second.

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My question is, what is the
height of this cliff?

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I think this is a good time
to actually introduce the

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direction notion of
velocity, to show

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you this scalar quantity.

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Let's assume up is positive,
and down is negative.

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My velocity is actually 100
meters per second down-- I

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could have assumed
the opposite.

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The final velocity is 100 meters
per second down, and

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since we're saying that down
is negative, and gravity is

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always pulling you down, we're
going to say that our

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acceleration is equal to
gravity, which is equal to

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minus 10 meters per
second squared.

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I just wrote that ahead of
times, because when we're

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dealing with anything of
throwing or jumping or

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anything on this planet, we
could just use this constant--

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the actual number is 9.81, but
I want to be able to do this

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without a calculator, so I'll
just stick with minus 10

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meters per second squared.

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It's pulling me down, so that's
why the minus is there.

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My question is: I know my
initial velocity, I know my

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final velocity, right before I
hit the ground or right when I

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hit the ground, what's
the distance?

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In this circumstance, what
does distance represent?

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Distance would be the height of
the cliff, and so how do we

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figure this out?

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What's the only formula that
we know for distance, or

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actually the change in distance,
but in this case,

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it's the same thing.

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Change in distance is equal
to the average velocity.

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When you learned this in middle
school, or probably

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even elementary school, you
didn't say average velocity,

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because you always assumed
velocity was constant-- the

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average and the instantaneous
velocity was

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kind of the same thing.

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Now, since the velocity is
changing, we're going to say

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the average velocity.

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So, the change in distance is
equal to the average velocity

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times time.

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This should be intuitive
to you at this point.

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Velocity really is just distance
divided by time, or

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actually, change in distance
divided by times change in

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time-- or, change in
distance divided by

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change in times is velocity.

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Let me actually change this
to change in time.

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Since we always assume-- or we
normally assume-- that we

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start at distance is equal to 0,
and we assume that start at

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time is equal to 0, we can write
distance is equal to

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velocity average times time.

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Maybe later on we'll do
situations where we're not

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starting at time 0, or distance
0, and in that case,

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we will have to be a little more
formal and say change in

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distance is equal to average
velocity the change in time.

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This is a formula we know,
and let's see what

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we can figure out.

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Can we figure out the
average velocity?

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The average velocity is just
the average of the initial

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velocity and the
final velocity.

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The average velocity is just
equal to the average of these

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two numbers: so, minus 100 plus
0 over 2-- and I'm just

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averaging the numbers-- equals
minus 50 meters per second.

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We were able to figure
that out, so can

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we figure out time?

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We know also that velocity,
or let's say the change in

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velocity, is equal to the
final velocity minus the

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initial velocity.

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This is nothing fancy-- you
don't have to memorize this.

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This hopefully is intuitive to
you, that the change is just

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the final velocity minus the
initial velocity, and that

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that equals acceleration
times time.

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So what's the change in velocity
in this situation?

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Final velocity is minus 100
meters per second, and then

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the initial velocity is 0, so
the change in velocity is

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equal to minus 100 meters
per second.

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I'm kind of jumping in and out
of the units, but I think you

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get what I'm doing.

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That equals acceleration
times time-- what's the

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acceleration?

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It's minus 10 meters per second
squared, because I'm

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going straight down-- minus 10
meters per second squared

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times time.

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This is a pretty straightforward
equation.

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Let's divide both sides by the
acceleration, by the minus 10

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meters per second squared, and
you'll get time is equal to--

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the negatives cancel out, as
they should, because negative

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time is difficult, we're
assuming positive time, and

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it's good we got a positive
time answer-- but the

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negatives cancel out
and we get time

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is equal to 10 seconds.

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There we have it: we figured out
time, we figured out the

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average velocity, and so now we
can figure out the height

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of the cliff.

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The distance is equal to the
average velocity minus 50

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meters per second times
10 seconds.

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The distance-- this is going to
be an interesting notion to

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you-- the distance it's going
to be minus 500 meters.

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This might not make a lot of
sense to you-- what does minus

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500 meters mean?

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This is actually right, because
this formula is

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actually the change
in distance.

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We said if we did it formally,
it would be

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the change in distance.

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So if we have a cliff-- let me
change colors with it-- and if

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we assume that we start at this
point right here, and

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this distance is equal to 0,
then the ground, if this cliff

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is 500 hundred meters high, your
final distance-- this is

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the initial distance-- your
final distance df is actually

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going to be at minus
500 hundred meters.

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We could have done it the other
way around: we could

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have said this is plus 500
meters, and then this is 0,

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but all that matters is really
the change in distance.

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We're saying from the top of the
cliff to the ground, the

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change in distance is
minus 500 meters.

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And minus, based on our
convention, we said minus is

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down, so the change is 500
meters down, and that's height

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of the cliff.

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That's pretty interesting.

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If you go to a 500 meter cliff--
500 is about 1,500

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feet-- so that's roughly the
size of maybe a very tall

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skyscraper, like the
World Trade Center

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or the Sears Tower.

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If you jump off of something
like that, assuming no air

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resistance, which is a big
assumption, or if you were to

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drop a penny-- because a penny
has very little air

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resistance-- if you were to drop
a penny off of the top of

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Sears Tower or a building like
that, at the bottom it will be

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going 100 meters per second.

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That's extremely fast, and
that's why you shouldn't be

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doing it, because that is fast
enough to kill somebody, and I

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don't want to give you any bad
ideas if you're a bad person.

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It's just interesting that
physics allows you to solve

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these types of problems.

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In the next presentation, I'm
just going to keep doing

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problems, and hopefully you'll
realize that everything really

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just boils down to average
velocity-- change in velocity

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is acceleration times time,
and change in distance is

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equal to change in time times
average velocity, which we all

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did just now.

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I'll see you in the
next presentation.

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