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We'll now use that equation we
just derived to go back and

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solve-- or at least address--
that same problem we were

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doing before, so let's write
that equation down again.

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Actually, let's write
the problem down.

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Lets say I have the cliff
again, and so my initial

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distance is 0, but it goes
down 500 meters.

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I'm not going to redraw the
cliff, because it takes a lot

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of space up on my limited
chalkboard.

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We know that the change
in distance is equal

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to minus 500 meters.

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I'm still going to use the
example where I don't just

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drop the ball, or the penny, or
whatever I'm throwing off

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the cliff, but I actually throw
it straight up, so it's

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going to go up and slow down
from gravity, and then it will

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go to 0 velocity, and start
accelerating downwards.

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You could even say decelerating

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in the other direction.

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The initial velocity, vi, is
equal to 30 meters per second,

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and of course, we know that the
acceleration is equal to

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minus 10 meters per second
squared-- it's because

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acceleration gravity is always
pulling downwards, or towards

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the center of our planet.

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If we wanted to figure out the
final velocity, we could have

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just used the formula, and we
did this in the last video--

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vf squared is equal to
vi squared plus 2ad.

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Now what I want to do is use the
formula that we learned in

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the very last video to figure
out-- how long does it take to

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get to the bottom, and
to hit the ground?

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Let's use that formula: we
derived that the change in

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distance is equal to the initial
velocity times time

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plus acceleration time
squared over 2, and

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that's initial velocity.

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The change in distance is minus
500, and that's equal to

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the initial velocity-- that's
positive, going upwards, 30

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meters per second, 30t.

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I'm not going to write the units
right now, because I'll

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run out of space, but you can
redo it with the units, and

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see that the units
do work out.

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When you square time, you have
to square the time units,

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although we're solving
for time.

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Then, plus acceleration, and
acceleration is minus 10, and

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we're going to divided it by 2,
so it's minus 5t squared.

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We have minus 500 is equal to
30t plus minus 5t, and we

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could just say minus
5t squared, and

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get rid of this plus.

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At first, you say, Sal-- there's
a t, that's t to the

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first, and t to the second,
how do I saw solve this?

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Hopefully, you've taken algebra
two or algebra one, in

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some places, and you remember
how to solve this.

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Otherwise, you're about to learn
the quadratic equation,

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although I recommend you go
back, and learn about

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factoring in the quadratic
equation, which there are

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videos on that I've
put on Youtube.

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I hope you watch those first
if you don't remember.

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We can do this-- let's put these
two right terms on the

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left hand side, and then we'll
use the quadratic equation to

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solve, because I don't think
this is easy to factor.

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We'll get 5t squared minus 30t
minus 500 is equal to 0-- I

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just took these terms and put
them on the left side.

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We could divide both sides by
5, just to simplify things,

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and so we get t squared minus
6t minus 100 is equal to 0.

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I could do that, because 0
divided by 5 is just five, so

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I just cleaned it
up a little bit.

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Let's use the quadratic
equation, and for those of us

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who need a refresher,
I'll write it down.

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The roots of any quadratic--
in this case, it's t we're

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solving for-- t will equal
negative b plus or minus the

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square root of b squared minus
4ac over 2a, where a is a

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coefficient on this term, b is
the coefficient on this term,

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negative 6, and c is the
constant, so minus 100.

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Let's just solve.

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We get t is equal to negative
b-- so negative this term.

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This term is negative 6, so if
we make it a negative, it

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becomes plus 6.

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It becomes 6 plus or minus the
square root of b squared, so

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it's minus 6 squared, 36,
minus 4 times a, and the

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coefficient on a is here,
and that's just times 1.

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With 4ac, c is a constant term,
minus 100, minus 4 times

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1 times minus 100, and all of
that is over 2a-- a is 1

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agains, so all of
that is over 2.

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That just equals 6 plus or minus
the square root-- this

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is minus 4 times minus 100, and
these become pluses, so it

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becomes 36 plus 400.

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So, 6 plus or minus
436 divided by 2.

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This is not a clean number,
and if you type into a

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calculator, it's something on
the order of about 20.9.

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We can just say approximately
21-- you might want to get the

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exact number, if you're actually
doing this on a test,

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or trying to send something to
Mars, but for our purposes, I

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think you'll get the point.

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I'll say it approximately now,
because we're going to be a

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little off, but just to have
clean numbers, this is

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approximately 21--
it's like 20.9.

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We'll say 6 plus or minus--
let me just write

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20.9-- 20.9 over 2.

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Let me ask you a question:
if I do 6 minus

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20.9, what do I get?

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I get a negative number,
and does a

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negative time make sense?

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No, it does not, and that means
that somehow in the

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past-- I don't want to get
philosophical-- the negative

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time in this context will
not make sense.

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Really, we can just stick to the
plus, because 6 minus 20

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is negative, so there's only one
time that will solve this

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in a meaningful way.

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Time is approximately equal to
6 plus 20.9, so that's 26.9

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over 2, and that equals
13.45 seconds.

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That's interesting.

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I think if you remember way
back, maybe four or five

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videos ago, when we first did
this problem, we just dropped

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the penny straight
from the height.

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Actually, in that problem, I
gave you the time-- I said it

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took 10 seconds to hit the
ground, and we worked

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backwards to figure out that the
cliff was 500 meters high.

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Now, if you're here at the top
of a 500 cliff, or building,

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and you drop something that has
air resistance-- like a

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penny, that has very air
resistance-- it would take 10

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seconds to reach the ground,
assuming all of our

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assumptions about gravity.

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But if you were to throw the
penny straight up, off the

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edge of the cliff, at 30 meters
per second right here,

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it's going to take 13.5--
roughly, 13 and 1/2 seconds--

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to reach the ground.

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It takes a little bit longer,
and that should make sense

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because-- I have time to
draw a little picture.

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In the first case, I just took
the penny, and its motion just

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went straight down.

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In the second case, I took the
penny-- it first went up, and

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then it went down.

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It had all the time when it went
up, and then it went down

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a longer distance, so it makes
sense that this time-- this

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was 10 seconds, while this time
was 13.45 five seconds.

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You can kind of say that it
took-- well, you actually

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can't say that.

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I don't want to get too
involved, but I hopefully this

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make sense to you.

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If you have a smaller number
here, you should have gone and

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checked your work, because why
would it take less time when I

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throw the object straight up?

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Hopefully, that gave you a
little bit more intuition, and

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you really do have in your
arsenal now all of the

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equations-- and hopefully, the
intuition you need-- to solve

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basic projectile problems.

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I'll now probably do a couple
more videos where I just do a

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bunch of problems, just to
really drive the points home.

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Then, I'll expand these
problems to two

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dimensions and angles.

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Before we get there, you might
want to refresh your

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trigonometry.

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I'll see you soon.

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