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What I want to do
in this video is

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tackle a problem that would
be considered pretty difficult

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for most first-year
physics students.

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And you, frankly,
probably wouldn't

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be expected to solve
a problem like this

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in most first-year
physics class.

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Or if you're in an
advanced or honors class

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you might be expected or it
might be a bonus problem.

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But it's an interesting
type of problem.

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Because what we're
going to do is,

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we're going to launch a
projectile on an incline.

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So maybe we're on
the side of the hill.

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So it's a hill.

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Let me do it in green.

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So let's say we're on the
side of the hill like this.

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And let's say that we know
the inclination of the hill.

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The hill's inclination is 30
degrees off the horizontal.

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So this is the horizontal
right over here.

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So that is the
inclination of the hill.

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And then we're going
to launch a projectile

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at 10 meters per second.

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We're going to launch it
at 10 meters per second.

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And the angle with the
hill is 15 degrees.

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So at a 15 degree
angle with the hill.

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And the reason why
this is more difficult

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than the traditional projectile
motion problems is, well, we

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could think about it.

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The projectile is
going to be launched

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and it is going to eventually
land at some point on the hill.

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But we can't do
the simple figure

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out how long it's in the air
using its vertical velocity

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because we don't know what
the vertical displacement

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for this thing is
going to be, unless we

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know how far down
the hill it lands.

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Because the further
down the hill

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it lands, the higher the
vertical displacement.

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So we have to think
about both the horizontal

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and the vertical displacement
at the same time.

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And as we walk
through this, you'll

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see how that can be done.

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So I guess the first thing
that we really would always

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want to do whenever
you want to try

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to solve this type of problem is
break up our velocity into both

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the horizontal and
vertical components.

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So the vertical
component of our velocity

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is going to be the magnitude
of our total velocity, 10

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meters per second, times--
and be very careful here--

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not the sine of 15 degrees,
but the sine of the angle

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with the horizontal.

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So times the sine of 45 degrees.

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And I go into it in a lot more
detail in previous videos.

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For the sake of time,
I won't go into it.

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But it really just
comes from sohcahtoa.

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If we were to draw the
vertical component,

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it would look like this.

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This is the angle.

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The sine of 45 degrees
is equal to the opposite

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over the hypotenuse or
the hypotenuse times

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the sine of 45 degrees is equal
to the vertical component.

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That's where it's coming from.

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Let me get rid of some
of the stuff I just

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drew just so it makes
it a little bit cleaner.

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And the horizontal component
of our velocity is going to be,

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by the same logic, 10
cosine of 45 degrees.

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Now let's think about what
the horizontal displacement

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is going to be.

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And I'm just going to go
straight to the formula

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that we've derived in
the last few videos.

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The horizontal
displacement is going

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to be our initial-- sorry.

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Let's do the vertical
displacement.

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The vertical
displacement-- I could

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have done the horizontal
displacement first--

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but the vertical
displacement is going

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to be our initial
vertical velocity, which

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we know as 10 sine
of 45 degrees.

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And by the way, we could
just solve that right now.

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What is the sine of 45 degrees?

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Sine of 45 degrees is
square root of 2/2.

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Cosine of 45 degrees is
also square root of 2/2.

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So both of these values,
10 times square root of 2/2

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is 5 square roots of 2.

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So this whole thing
right over here

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is 5 square roots of
2 meters per second.

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That's our vertical velocity.

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And our horizontal
velocity is also

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5 square roots of 2
meters per second.

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So that simplifies
things a little bit.

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But anyway, we
were talking about

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our vertical displacement.

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Our vertical
displacement is going

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to be our initial vertical
velocity, 5 square roots of 2,

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times our change in time
plus the acceleration.

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Well, we know what
the acceleration is.

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It's negative 9.8 meters
per second squared.

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So let me write minus 9.8.

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I'm not going to write the
units here so that we say space.

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Times our change
in time squared.

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All of that over 2.

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We derive this in several
videos, especially the last few

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where we do these two
dimensional projectile motions.

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So this gives us
our displacement

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in the y direction.

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And I can simplify
this a little bit.

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Our displacement in
the vertical direction

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is equal to 5 square
roots of 2 times

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delta t times-- let me do it in
the same-- times delta t change

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in time minus 4.9 times
our change in time squared.

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So we know we have this
constraint right over here.

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So this gives us our
vertical displacement

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as a function of time.

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Let's think about our
horizontal displacement

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as a function of time.

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Our horizontal
displacement is going

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to be equal to our
horizontal velocity, which

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is 5 square roots of 2
times our change in time.

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Now what can we do next?

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Well, we have to have
some relationship

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between our horizontal
displacement

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and our vertical displacement.

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And that relationship
is going to be

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given to us by this incline.

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So wherever we
land-- let's say this

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is where we eventually
do end up landing.

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Well, let's think
about our horizontal

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and our vertical
displacements and what

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their relationship
would have to be.

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So if this is where
we land, then this

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would be our-- let me do it
in the same colors-- that

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right there would be my
vertical displacement.

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I would move that far up.

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And then our
horizontal displacement

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will be this right over
here, will be that length

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right over there.

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So that is our
horizontal displacement.

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So what is the
relationship between

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our vertical displacement and
our horizontal displacement?

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And we know that this angle
right over here is 30 degrees.

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So we can use some
basic trigonometry.

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We have a right triangle.

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We know the opposite
side from the angle.

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We know the adjacent side.

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And the trig function that uses
the opposite and the adjacent

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is the tangent function.

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So we get the
tangent of 30 degrees

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is going to be equal
to the magnitude

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of our vertical displacement
over the magnitude

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of our horizontal displacement.

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And the tangent of
30 degrees, that's

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the same thing as
the sine of 30 over--

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let me just do it over here.

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So the tangent of 30
degrees is the same thing

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as the sine of 30 degrees
over the cosine of 30 degrees.

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Let me do this a
little bit neater.

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And the sine of
30 degrees is 1/2.

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This is equal to 1/2.

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And the cosine of 30 degrees
is the square root of 3/2.

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So this is equal to 1/2 times
2 over the square root of 3,

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which is equal to 1 over
the square root of 3.

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So we get the magnitude
of our vertical component

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over the magnitude of our
horizontal component--

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is our horizontal
component right over here--

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is equal to 1 over
the square root of 3.

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What's useful about
this is it gives us

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a relationship
between our horizontal

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and vertical component,
or between our vertical

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and our horizontal components.

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And we can use this
constraint right over here

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to then solve for
one of these two.

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And let me show you
how we'll do it.

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So let's just
explicitly write this.

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Well, let's do it this way.

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Let us cross multiply
here, which is really

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the same thing as multiplying
both sides by the square root

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of 3 and the magnitude of
our horizontal component.

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We get the square root
of 3 times the magnitude.

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And both of these are
going to be positive.

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Well, let me just
write it this way.

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Times the magnitude of
our vertical component

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is going to be equal
to the magnitude

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of our horizontal component.

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So right, just like that.

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So we now have a relationship
between the length of these two

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vectors.

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And we can use this
relationship to substitute back

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into the constraints
that we already have.

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So the second constraint
right over here,

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let's use this information.

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The second constraint says
that our horizontal component

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00:09:17,190 --> 00:09:19,600
of our displacement is
equal to 5 square roots

200
00:09:19,600 --> 00:09:21,810
of 2 times our change in time.

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00:09:21,810 --> 00:09:24,120
Or another way of
thinking about it,

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00:09:24,120 --> 00:09:26,960
if we divide both sides
by 5 square roots of 2,

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00:09:26,960 --> 00:09:29,930
we get our change
in time is equal

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00:09:29,930 --> 00:09:39,260
to the horizontal component
of our displacement divided

205
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by 5 square roots of 2.

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00:09:41,430 --> 00:09:43,590
But we also know that
the horizontal component

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00:09:43,590 --> 00:09:45,240
of our displacement
is the square root

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00:09:45,240 --> 00:09:49,520
of 3 times the vertical
component of our displacement.

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00:09:49,520 --> 00:09:52,440
Here I explicitly wrote
the magnitude notation.

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00:09:52,440 --> 00:09:55,460
When we start dealing with
either just the vertical

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00:09:55,460 --> 00:09:58,000
or the horizontal
component, I can just

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00:09:58,000 --> 00:09:59,940
write it like this,
because it's either

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00:09:59,940 --> 00:10:01,970
going to be a positive
or a negative value.

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00:10:01,970 --> 00:10:04,580
And that specifies both the
magnitude and the direction.

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00:10:04,580 --> 00:10:06,830
So what I'm going to do right
over here, and obviously

216
00:10:06,830 --> 00:10:08,996
the way I've drawn it right
over here, both of these

217
00:10:08,996 --> 00:10:10,370
are going to be positive values.

218
00:10:10,370 --> 00:10:12,536
It's upwards displacement
in the vertical direction,

219
00:10:12,536 --> 00:10:14,330
so that's positive
by our convention.

220
00:10:14,330 --> 00:10:15,620
And we're moving to the right.

221
00:10:15,620 --> 00:10:19,120
So that's positive,
also, by our convention.

222
00:10:19,120 --> 00:10:24,050
So I can rewrite this
over here as being

223
00:10:24,050 --> 00:10:32,500
equal to square root of 3 times
our vertical displacement.

224
00:10:32,500 --> 00:10:35,040


225
00:10:35,040 --> 00:10:37,220
And all of that's over
5 square roots of 2.

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00:10:37,220 --> 00:10:39,730


227
00:10:39,730 --> 00:10:41,910
Now, the whole
reason why I did this

228
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is, this expression right here
contains this information,

229
00:10:47,110 --> 00:10:50,060
contains the ratio between
our vertical displacement

230
00:10:50,060 --> 00:10:51,590
and our horizontal displacement.

231
00:10:51,590 --> 00:10:54,360
And it also contains
the information of,

232
00:10:54,360 --> 00:10:57,730
how does the horizontal
displacement, how does that

233
00:10:57,730 --> 00:10:59,580
change as a function of time?

234
00:10:59,580 --> 00:11:04,190
So our time needs
to be equal to this.

235
00:11:04,190 --> 00:11:06,060
So this is our
time as a function

236
00:11:06,060 --> 00:11:07,700
of our vertical
displacement now,

237
00:11:07,700 --> 00:11:10,440
not time as a function of
our horizontal displacement.

238
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And what we can do is, we
can use this constraint

239
00:11:12,820 --> 00:11:17,510
with our original
vertical displacement

240
00:11:17,510 --> 00:11:19,680
as a function of
time to then solve

241
00:11:19,680 --> 00:11:21,380
for our vertical displacements.

242
00:11:21,380 --> 00:11:22,150
So let's do that.

243
00:11:22,150 --> 00:11:24,530
Let's substitute this
business right here

244
00:11:24,530 --> 00:11:29,190
for delta t in this top
equation right over here.

245
00:11:29,190 --> 00:11:31,630
So if we do that-- and
I'll write it big--

246
00:11:31,630 --> 00:11:38,980
we get our vertical
displacement, right over there,

247
00:11:38,980 --> 00:11:41,880
is equal to 5 square
roots of 2 times delta t.

248
00:11:41,880 --> 00:11:44,140
So it's 5 square roots of 2.

249
00:11:44,140 --> 00:11:46,330
Delta t is all of this
business over here.

250
00:11:46,330 --> 00:11:48,480
So 5 square roots
of 2 times delta t.

251
00:11:48,480 --> 00:11:52,936
Delta t is square root of 3
times our vertical component.

252
00:11:52,936 --> 00:11:54,310
All of that, really
the magnitude

253
00:11:54,310 --> 00:11:58,510
of our vertical component,
over 5 square roots of 2.

254
00:11:58,510 --> 00:12:00,579
So that's that right there.

255
00:12:00,579 --> 00:12:02,870
Or actually, we could look
at this one right over here.

256
00:12:02,870 --> 00:12:04,078
We could use this constraint.

257
00:12:04,078 --> 00:12:05,430
This is just simplified.

258
00:12:05,430 --> 00:12:11,150
So then we have minus 4.9
times delta t squared.

259
00:12:11,150 --> 00:12:13,700
So delta t squared is
this quantity squared.

260
00:12:13,700 --> 00:12:14,710
I'll just write it out.

261
00:12:14,710 --> 00:12:16,310
I don't want to
skip too many steps.

262
00:12:16,310 --> 00:12:19,060
So delta t once again is
the square root of 3 times

263
00:12:19,060 --> 00:12:21,100
the vertical component.

264
00:12:21,100 --> 00:12:25,620
All of that over 5 square
roots of 2 squared.

265
00:12:25,620 --> 00:12:27,510
And now what does this give us?

266
00:12:27,510 --> 00:12:30,050
So now we literally have
a quadratic equation

267
00:12:30,050 --> 00:12:31,530
with only one variable.

268
00:12:31,530 --> 00:12:33,050
So we can solve for this.

269
00:12:33,050 --> 00:12:34,120
But let me rewrite it.

270
00:12:34,120 --> 00:12:35,060
Let me simplify it.

271
00:12:35,060 --> 00:12:36,920
So we have our
vertical component

272
00:12:36,920 --> 00:12:40,249
is equal to-- now we have
a 5 square roots of 2

273
00:12:40,249 --> 00:12:42,040
in the numerator and
one in the denominator

274
00:12:42,040 --> 00:12:43,270
and they cancel out.

275
00:12:43,270 --> 00:12:48,040
So we get the square root of 3
times our vertical component,

276
00:12:48,040 --> 00:12:50,010
the magnitude of our
vertical component.

277
00:12:50,010 --> 00:12:52,540
It's actually also specifying,
well, the magnitude,

278
00:12:52,540 --> 00:12:57,180
we can say, for now, although
I'm not using that notation.

279
00:12:57,180 --> 00:13:01,235
So then we have minus 4.9
times this quantity squared.

280
00:13:01,235 --> 00:13:03,360
So that's going to be, the
square root of 3 squared

281
00:13:03,360 --> 00:13:10,340
is 3 times the vertical
component squared.

282
00:13:10,340 --> 00:13:14,561
And then over 5 squared,
which is 25, times 2.

283
00:13:14,561 --> 00:13:16,060
That's the square
root of 2 squares.

284
00:13:16,060 --> 00:13:19,230
So 25 times 2 is 50.

285
00:13:19,230 --> 00:13:22,490
And so we get, if we were to
simplify this a little bit

286
00:13:22,490 --> 00:13:22,990
more.

287
00:13:22,990 --> 00:13:25,710
Let's subtract this
from both sides.

288
00:13:25,710 --> 00:13:28,140
We get-- I'll do it
all in one color-- 0

289
00:13:28,140 --> 00:13:35,320
is equal to the square
root of 3 minus 1

290
00:13:35,320 --> 00:13:39,790
times our vertical component.

291
00:13:39,790 --> 00:13:41,880
Because if we subtract
this from both sides,

292
00:13:41,880 --> 00:13:44,520
that's square root of 3
times our vertical component

293
00:13:44,520 --> 00:13:46,327
minus 1 times our
vertical component.

294
00:13:46,327 --> 00:13:48,160
So that's the square
root of 3 minus 1 times

295
00:13:48,160 --> 00:13:49,570
our vertical component.

296
00:13:49,570 --> 00:13:51,290
And then we have all
of this business.

297
00:13:51,290 --> 00:14:01,800
Minus 4.9 times 3 over 50 times
our vertical component squared.

298
00:14:01,800 --> 00:14:03,540
And lucky for us, we
can just factor out

299
00:14:03,540 --> 00:14:08,080
one of these s sub y's over
here, one of these vectors.

300
00:14:08,080 --> 00:14:11,040
And so we can get-- and I'll
just do that in place so that I

301
00:14:11,040 --> 00:14:13,540
don't-- well, let me just-- I
don't want to skip many steps.

302
00:14:13,540 --> 00:14:20,080
So this is equal to the square
root of 3 minus 1, minus 4.9.

303
00:14:20,080 --> 00:14:22,410
Let me do it in that color.

304
00:14:22,410 --> 00:14:25,370
So let me do it like this.

305
00:14:25,370 --> 00:14:32,360
It's equal to the square root
of 3 minus 1 minus 4.9 times

306
00:14:32,360 --> 00:14:38,910
3 over 50 times one of these,
times our vertical component.

307
00:14:38,910 --> 00:14:42,110
And then we factored one of
those vertical components out.

308
00:14:42,110 --> 00:14:44,160
So we factored one of them out.

309
00:14:44,160 --> 00:14:47,150
So the vertical component
of our displacement

310
00:14:47,150 --> 00:14:50,350
could either be-- so we have
the product of two things that

311
00:14:50,350 --> 00:14:51,404
equal 0.

312
00:14:51,404 --> 00:14:52,820
So our vertical
displacement could

313
00:14:52,820 --> 00:14:56,230
be 0, which is true, because
at some point in the path

314
00:14:56,230 --> 00:14:58,450
we literally had 0
vertical displacement.

315
00:14:58,450 --> 00:14:59,950
That was literally
where we started.

316
00:14:59,950 --> 00:15:01,280
But that's not the
answer we're looking for.

317
00:15:01,280 --> 00:15:03,480
We're looking for this
vertical displacement.

318
00:15:03,480 --> 00:15:05,050
So either this is going to be 0.

319
00:15:05,050 --> 00:15:06,841
But that's just kind
of the obvious answer.

320
00:15:06,841 --> 00:15:10,330
Or all of this business
is going to be equal to 0.

321
00:15:10,330 --> 00:15:13,520
But this is pretty easy
to solve for 0 over here.

322
00:15:13,520 --> 00:15:20,412
So we get square root
of 3 minus 1 minus 4.9.

323
00:15:20,412 --> 00:15:22,120
Let me just calculate
all of these things

324
00:15:22,120 --> 00:15:24,490
just so that I don't have
to keep writing them.

325
00:15:24,490 --> 00:15:35,570
So we get the square root of
3 minus 1 is equal to 0.73205.

326
00:15:35,570 --> 00:15:37,920
So I'll just write 0.732 here.

327
00:15:37,920 --> 00:15:41,990
So this is equal to 0.732.

328
00:15:41,990 --> 00:15:43,810
That's that part
right over there.

329
00:15:43,810 --> 00:15:47,370
And then 4.94.

330
00:15:47,370 --> 00:15:49,629
I know this problem
is getting long.

331
00:15:49,629 --> 00:15:51,920
Just pause it and take a
break if you're getting tired.

332
00:15:51,920 --> 00:15:59,170
4.9 times 3 divided by
50 is equal to 0.294.

333
00:15:59,170 --> 00:16:03,580
So minus 0.294.

334
00:16:03,580 --> 00:16:06,710
I could put a 0 out front just
to make it clear where we are.

335
00:16:06,710 --> 00:16:10,820
Times this, times our
vertical component.

336
00:16:10,820 --> 00:16:12,650
This could also be equal to 0.

337
00:16:12,650 --> 00:16:14,480
Either this is 0 or that is 0.

338
00:16:14,480 --> 00:16:16,490
When this is 0, it gives
us the obvious answer.

339
00:16:16,490 --> 00:16:17,970
We're more interested in this.

340
00:16:17,970 --> 00:16:22,610
To solve for this, we can
add this to both sides.

341
00:16:22,610 --> 00:16:30,280
And we get 0.732 is
equal to negative 0.294

342
00:16:30,280 --> 00:16:32,430
times the vertical component.

343
00:16:32,430 --> 00:16:34,380
We are in the home stretch.

344
00:16:34,380 --> 00:16:42,261
We divide both sides by
this, by negative 0.29.

345
00:16:42,261 --> 00:16:43,760
Oh sorry, this will
now be positive.

346
00:16:43,760 --> 00:16:47,520
I almost made a careless mistake
after 16 minutes of video.

347
00:16:47,520 --> 00:16:54,590
So now we divide
both sides by 0.294

348
00:16:54,590 --> 00:16:58,130
and we get our
vertical displacement.

349
00:16:58,130 --> 00:17:00,480
So this, I think, a drum
roll might be in order.

350
00:17:00,480 --> 00:17:04,530
So we have 0.732 and
all of this business,

351
00:17:04,530 --> 00:17:10,000
but I'll just round it there,
divided by 0.294 gives us

352
00:17:10,000 --> 00:17:15,280
a vertical displacement of,
if we round, 2.50 meters.

353
00:17:15,280 --> 00:17:18,160
Or 2.49 meters, I should say.

354
00:17:18,160 --> 00:17:23,800
So this is equal to 2.49 meters.

355
00:17:23,800 --> 00:17:25,380
This is exciting.

356
00:17:25,380 --> 00:17:29,070
This is equal to 2.49 meters.

357
00:17:29,070 --> 00:17:31,280
And now we can figure out
the horizontal displacement

358
00:17:31,280 --> 00:17:33,260
pretty easily,
because we know that

359
00:17:33,260 --> 00:17:35,870
the horizontal displacement
is square root of 3 times

360
00:17:35,870 --> 00:17:37,639
the vertical displacement.

361
00:17:37,639 --> 00:17:38,680
So let's figure that out.

362
00:17:38,680 --> 00:17:40,370
That's the vertical
displacement.

363
00:17:40,370 --> 00:17:43,510
Let's multiply that times
the square root of 3.

364
00:17:43,510 --> 00:17:46,550
And we get 4.31 meters.

365
00:17:46,550 --> 00:17:49,580
So we get the
horizontal displacement.

366
00:17:49,580 --> 00:17:54,580
The horizontal displacement
is equal to 4.31 meters.

367
00:17:54,580 --> 00:17:59,080
So this is equal to 4.31 meters.

368
00:17:59,080 --> 00:18:01,280
So we actually now know
the total displacement

369
00:18:01,280 --> 00:18:03,834
in the vertical direction and
in the horizontal direction.

370
00:18:03,834 --> 00:18:05,000
And I'll leave it up to you.

371
00:18:05,000 --> 00:18:07,930
If you wanted to figure out
exactly how far along the hill

372
00:18:07,930 --> 00:18:10,010
we traveled, you
can just use both

373
00:18:10,010 --> 00:18:12,130
of these values in the
Pythagorean theorem

374
00:18:12,130 --> 00:18:13,960
to essentially figure
out the hypotenuse

375
00:18:13,960 --> 00:18:16,061
of this right triangle.

376
00:18:16,061 --> 00:00:00,000