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Let's say we're going to
shoot some object into

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the air at an angle.

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Let's say its speed is s and
the angle at which we shoot

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it, the angle above the
horizontal is theta.

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What I want to do in this video
is figure out how far

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this object is going to travel
as a function of the angle and

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as a function of the speed, but
we're going to assume that

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we're given the speed.

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That that's a bit
of a constant.

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So if this is the ground right
here, we want to figure out

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how far this thing is
going to travel.

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So you can imagine, it's going
to travel in this parabolic

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path and land at some
point out there.

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And so if this is at distance 0,
we could call this distance

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out here distance d.

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Now whenever you do any problem
like this where you're

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shooting something off at an
angle, the best first step is

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to break down that vector.

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Remember, a vector is something
that has magnitude

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and direction.

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The magnitude is s.

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Maybe feet per second
or miles per hour.

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And the direction is theta.

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So if you have s and theta,
you're giving me a vector.

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And so what you want to do is
you want to break this vector

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down into its vertical and
horizontal components first

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and then deal with
them separately.

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One, to help you figure out how
long you're in the air.

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And then, the other to
figure out how far

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you actually travel.

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So let me make a big version
of the vector right there.

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Once again, the magnitude
of the vector is s.

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So you could imagine that the
length of this arrow is s.

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And this angle right
here is theta.

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And to break it down into its
horizontal and vertical

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components, we just set up a
right triangle and just use

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our basic trig ratios.

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So let me do that.

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So this is the ground
right there.

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I can drop a vertical from the
tip of that arrow to set up a

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right triangle.

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And the length of the-- or the
magnitude of the vertical

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component of our velocity
is going to be this

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length right here.

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That is going to be-- you could
imagine, the length of

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that is going to be our
vertical speed.

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So this is our vertical speed.

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Maybe I'll just call that
the speed sub vertical.

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And then, this right here, the
length of this part of the

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triangle-- let me do that
in a different color.

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The length of this part of the
triangle is going to be our

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horizontal speed, or the
component of this velocity in

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the horizontal direction.

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And I use this word velocity
when I specify

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a speed and a direction.

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Speed is just the magnitude
of the velocity.

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So the magnitude of this side
is going to be speed

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horizontal.

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And to figure it out,
you literally use

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our basic trig ratios.

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So we have a right triangle.

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This is the hypotenuse.

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And we could write down
soh cah toa up here.

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Let me write it down
in yellow.

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soh cah toa.

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And this tells us that sine is
opposite over hypotenuse,

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cosine is adjacent over
hypotenuse and tangent is

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opposite over adjacent.

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So let's see what we can do.

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We're assuming we know
theta, we know s.

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We want to figure out what the
vertical and the horizontal

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components are.

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So what's the vertical component
going to be?

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Well the vertical component
is opposite this theta.

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But we know the hypotenuse is
s, so we could use sine

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because that deals with the
opposite and the hypotenuse.

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And the sine function tells us
that sine of theta-- actually,

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let me do this in green since
we're doing all the vertical

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stuff in green.

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Sine of theta is going to be
equal to opposite, which is

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the magnitude of our
vertical velocity.

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So the opposite side is this
side right here, over our

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hypotenuse.

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And our hypotenuse
is the speed s.

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And so if we want to solve for
our vertical velocity or the

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vertical component of our
velocity, we multiply both

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sides of this equation by s.

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So you get s sine of theta
is equal to the vertical

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component of our velocity,
s sine of theta.

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And now for the horizontal
component we do the same

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thing, but we don't
use sine anymore.

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This is now adjacent
to the angle.

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So cosine deals with the
adjacent side and the

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hypotenuse.

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So we could say that the cosine
of theta is equal to

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the adjacent side to the angle,
that is the horizontal

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speed, over the hypotenuse.

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The hypotenuse is this length
right here, over s.

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So if we want to solve for the
horizontal speed or the

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horizontal component or the
magnitude of the horizontal

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component, we'd just multiply
both sides times s.

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And you get s cosine of
theta is equal to

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the horizontal component.

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So we now know how fast we are
travelling in this direction,

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in the horizontal component.

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We know that that is going
to be s cosine of theta.

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And we know in the vertical
direction-- let me do that in

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the vertical direction, the
magnitude is s sine of theta.

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It is s sine of theta.

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So now that we've broken up into
the two components, we're

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ready to figure out how long
we're going to be in the air.

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