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Welcome back.

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Let's continue doing projectile
motion problems. I

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think this video will be
especially entertaining,

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because I will teach you a game
that you can play with a

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friend, and it's called let's
see how fast and how high I

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can throw this ball.

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You'd be surprised--
it's actually quite

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a stimulating game.

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Let me just write
down everything

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we've learned so far.

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Change in distance is
equal to the average

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velocity times time.

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We know that change in
velocity is equal to

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acceleration times time.

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We also know that average
velocity is equal to the final

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velocity plus the initial
velocity over 2.

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We know the change in velocity,
of course, is equal

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to the final velocity minus
the initial velocity.

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This should hopefully be
intuitive to you, because it's

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just how fast you're going at
the end, minus how fast you're

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going at the beginning,
divided-- oh no, no division,

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it's just that I got
stuck in a pattern.

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It's just vf minus
vi, of course.

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You probably already knew this
before you even stumbled upon

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my videos, but-- the two
non-intuitive ones that we've

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learned, they're really just
derived from what I've just

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written up here.

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If you ever forget them, you
should try to derive them.

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Actually, you should try to
derive them, even if you don't

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forget them, so that you
when you do forget it,

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you can derive it.

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It's change in distance-- let
me change it to lowercase d,

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just to confuse you-- is equal
to the initial velocity times

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time plus at squared over 2,
and that's one of what I'll

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call the non-intuitive
formulas.

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The other one is the final
velocity squared is equal to

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the initial velocity
squared plus 2ad.

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We've derived all of these,
and I encourage you to try

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rederive them.

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But using these two formulas
you can play my fun game of

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how fast and how high did
I throw this ball?

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All you need is your arm, a
ball, a stopwatch, and maybe

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some friends to watch
you throw the ball.

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So how do we play this game?

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We take a ball, and we throw
it as high as we can.

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We see how long does the
ball stay in the air?

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What do we know?

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We know the time for the ball
to leave your hand, to

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essentially leave the ground and
come back to the ground.

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We are given time, and
what else do we know?

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We know acceleration-- we know
acceleration is this minus 10

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meters per second.

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If you're actually playing
this game for money, or

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something, you would probably
want to use a more accurate

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acceleration-- you could look
it up on Wikipedia.

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It's minus 9.81 meters
per second squared.

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Do we know the change
in distance?

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At first, you're saying-- Sal,
I don't know how high this

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ball went, but we're talking
about the change in distance

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over the entire time.

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It starts at the ground--
essentially at the ground,

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because I'm assuming that you're
not 100 feet tall, and

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so you're essentially at the
ground-- so it starts at the

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ground, and it ends of the
ground, so the change in

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distance of delta d is 0.

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It starts with at the ground
and ends at the ground.

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This is interesting-- this is a
vector quantity, because the

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direction matters.

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If I told you how far did the
ball travel, then you'd have

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to look at its path, and say
how high did it go, and how

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high would it come back?

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Actually, if you want to be
really exact, the change in

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distance would be the height
from your hand when the ball

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left your hand, to the ground--
so, if you're 6 feet

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tall, or 2 meters tall, the
change in distance would

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actually be minus 2 meters,
but we're not going to do

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that, because that would just be
too much, but you could do

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it if there's ever a close call
between you and a friend,

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and you're betting for money.

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You're given these things,
and we want to figure

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out a couple of things.

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The first thing I want to figure
out is how fast did I

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throw the ball, because that's
what's interesting-- that

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would be a pure test
of testosterone.

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How fast?

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I want to figure out vi--
vi equals question mark.

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Which of these formulas
can be used?

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Actually, I'm going do it first
with the formulas, and

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then I'm going to show you
almost an easier way to do it,

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where it's more intuitive.

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I want to show you that these
formulas can be used for fun

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with your friends.

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We know time, we know
acceleration, we know the

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change in distance, so we could
just solve for vi--

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let's do that.

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In this situation, change in
distance is 0-- let me change

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colors again just to change
colors-- so change in distance

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is 0 is equal to
vi times time.

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Let me put the g in for here,
so it's minus 10 meters per

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second squared divided by 2,
and it's minus 5 meters per

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second squared-- so it's
minus 5t squared.

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All I did it is that I took
minus 10 meters per second

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squared for a, divided it
by 2, and that's how I

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got the minus 5.

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If you used 9.81 or whatever,
this would be

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4.905 something something.

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Anyway, let's get back
to the problem.

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If we wanted to solve this
equation for vi,

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what could we do?

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This is pretty interesting,
because we

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could factor a t out.

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What's cool about these physics
equations is that

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everything we do actually has
kind of a real reasoning

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behind it in the real world,
so let me actually flip the

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sides, and factor out a t, just
to make it confusing.

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I get t times vi minus
5t is equal to 0.

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All I did is that I factored out
a t, and I could do this--

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I didn't have to use a quadratic
equation, or factor,

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because there wasn't a
constant term here.

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So I have this expression, and
if I were to solve it,

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assuming that you know vi is
some positive number, I know

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that there's two times where
this equation is true.

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Either t equals 0, or this term
equals 0-- vi minus 5t is

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equal to 0, or since I'm solving
for velocity, we know

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that vi is equal to 5t.

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That's interesting.

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So what does this say?

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If we knew the velocity, we
could solve it the other way.

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We could say that t is equal to
vi divided by 5-- these are

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the same things, just solving
for a different variable.

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But that's cool, because there
are two times when the change

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in distance is zero-- at time
equals 0, of course, the

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change in distance is zero,
because I haven't thrown the

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ball yet, and then, at a later
time, or my initial velocity

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divided by 5, it'll also
hit the ground again.

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Those are the two times
that the change

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in distance is zero.

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That's pretty cool.

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This isn't just math--
everything we're doing in math

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has kind of an application
in the real world.

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We've solved our equation--
vi is equal to 5t.

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So, if you and a friend go
outside and throw a ball, and

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you try to throw it straight
up--0 although we'll learn

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when we do the two dimensional
projectile motion that it

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actually doesn't matter if you
have a little bit of an angle

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on it, because the vertical
motion and the horizontal

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motions are actually
independent, or can be viewed

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as independent from each other--
this velocity you're

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going to get if you play this
game is going to be just the

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component of your velocity
that goes straight up.

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I know that might be a little
confusing, and hopefully it

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will make a little more sense in
a couple of videos from now

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when I teach you vectors.

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If you were to throw this ball
straight up, and time when it

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hits the ground, then this
velocity would literally the

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speed-- actually, the
velocity-- at which

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you throw the ball.

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So what would it be?

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If I threw a ball, and it took
two seconds to go up hit the

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ground, then I could
use this formula.

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This is actually 5
meters per second

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squared times t seconds.

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If it took 2 seconds-- so if
t is equal to 2-- then my

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initial velocity is equal
to 10 meters per second.

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You could convert that to
miles per hour-- we've

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actually done that in
previous videos.

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If you throw a ball that stays
up in the air for 10 seconds,

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then you threw it at 50 meters
per second, which is

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extremely, extremely fast.
Hopefully, I've taught you a

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little bit about a fun game.

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In the next video, I'll show
you how to figure out-- how

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high did the ball go?

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I'll see you soon.

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