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In the last video, I dropped
myself or a penny from the top

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of a cliff.

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We started off at 0 velocity,
obviously because it was

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stationary, and at the bottom,
it was 100 meters per second.

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We used that to figure out how
high the cliff was, and we

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figured out that the cliff
was 500 meters high.

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What I want to do now is let's
do that same problem, but

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let's do in a general form, and
see if we can figure out a

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general formula for a
problem like that.

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Let's say that you have the same
thing, and let's say the

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initial velocity-- you're given
the initial velocity,

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you're given the final velocity,
you're given the

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acceleration, and you want to
figure out the distance.

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This is what you're given,
and you want to

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figure out the distance.

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Doing it the exact same way we
did in that last presentation,

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but now we're now [INAUDIBLE]

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formulas, we know that the
change in distance is equal to

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the average velocity times--
we could actually say the

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change in time, but I'll just
say it with time, because we

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always assume that we
start with time

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equals 0-- times time.

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We know that the average
velocity is the final velocity

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plus the initial velocity
divided by 2, so that's the

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average velocity-- let me
highlight-- this is the same

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thing as this, and then
that times time.

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What's the time?

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You could figure out the time
by saying, we know how fast

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we're accelerating, and we know
the initial and final

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velocity, so we can figure out
how long we had to accelerate

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that fast to get that
change in velocity.

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Another way of saying that, or
probably a simpler way of

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saying that, is change in
velocity, which is the same

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thing as the final velocity
minus the initial velocity is

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equal to acceleration
times time.

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If you want to solve for time,
you could say the time-- if I

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just divide both sides of this
equation by a-- is equal to vf

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minus vi divided by a.

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We could take that and
substitute that into this

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equation, and remember-- this
is all change in distance.

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We say change in distance is
equal to-- let me write this

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term in yellow-- vf
plus vi over 2.

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Let me write this
term in green.

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That's times vf minus
vi over a.

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Then if we do a little
multiplying of expressions on

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the top-- you might have
recognized this-- this would

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be vf squared minus vi squared,
and then we multiply

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the denominators over 2a.

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So the change in distance is
equal to vf squared minus vi

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squared over 2a.

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That's exciting-- let me
write that over again.

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The change in distance is equal
to vf squared minus vi

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squared divided by 2
times acceleration.

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We could play around with this a
little bit, and if we assume

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that we started distance is
equal to 0, we could write d

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here, and that might
simplify things.

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If we multiply both sides by
2a, we get-- and I'm just

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going to switch this to
distance, if we assume that we

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always start at distances
equal to 0.

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di, or initial distance,
is always at point 0.

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We could right 2ad-- I'm just
multiplying both sides by 2a--

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is equal to vf squared minus vi
squared, or you could write

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it as vf squared is equal
to vi squared plus 2ad.

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I don't know what your physics
teacher might show you or

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written in your physics book,
but of these variations will

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show up in your physics book.

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The reason why I wanted to show
you that previous problem

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first is that I wanted to show
you that you could actually

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figure out these problems
without having to always

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memorize formulas and resort
to the formula.

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With that said, it's probably
not bad idea to memorize some

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form of this formula, although
you should understand how it

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was derived, and when
to apply it.

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Now that you have memorized it,
or I showed you that maybe

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you don't have to memorize
it, let's use this.

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Let's say I have the
same cliff, and it

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has now turned purple.

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It was 500 meters high-- it's
a 500 meter high cliff.

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This time, with the penny,
instead of just dropping it

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straight down, I'm going to
throw it straight up at

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positive 30 meters per second.

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The positive matters, because
remember, we said negative is

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down, positive is up-- that's
just the convention we use.

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Let's use this formula, or any
version of this formula, to

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figure out what our final
velocity was when we hit the

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bottom of the ground.

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This is probably the easiest
formula to use, because it

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actually solves for
final velocity.

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We can say the final velocity
vf squared is equal to the

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initial velocity squared-- so
what's our initial velocity?

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It's plus 30 meters per second,
so it's 30 meters per

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second squared plus 2ad.

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So, 2a is the acceleration of
gravity, which is minus 10,

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because it's going down, so it's
2a times minus 10-- I'm

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going to give up the units for
a second, just so I don't run

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out of space-- 2 times minus
10, and what's the height?

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What's the change in distance?

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Actually, I should be correct
about using change in

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distance, because it matters
for this problem.

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In this case, the final distance
is equal to minus

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500, and the initial distance
is equal to 0.

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The change in distance
is minus 500.

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So what does this get us?

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We get vf squared is equal to
900, and the negatives cancel

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out-- 10 times 500 is 5,000, and
5,000 times 2 is 10,000.

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So vf squared is equal
to 10,900.

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So the final velocity is equal
to the square root of 10,900.

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What is that?

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Let me bring over my trusty
Windows-provided default

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calculator.

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It's 10,900, and the
square root.

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It's about 104 meters per
second, so my final velocity

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is approximately-- that
squiggly equals is

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approximately-- 104
meters per second.

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That's interesting.

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If I just dropped something--
if I just drop it straight

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from the top-- we figured out
in the last problem that at

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the end, I'm at 100
meters per second.

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But this time, if I throw it
straight up at 30 meters per

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second, when the penny hits
the ground, it's actually

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going even faster.

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You might want to think about
why that is, and you might

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realize it.

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When I throw it up, the highest
point of the penny--

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if I throw it up at 30 meters
per second, the highest point

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of the penny is going to be
higher than 500 meters-- is

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going to make some positive
distance first, and then it's

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going to come down, so it's
going to have even more time

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to accelerate.

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I think that makes some
intuitive sense to you.

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That's all the time I have
now, and in the next

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presentation, maybe I'll use
this formula to solve a couple

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of other types of problems.

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