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## Projectiles launched at an angle

Current time:0:00Total duration:5:50

# Optimal angle for a projectile part 1: Components of initial velocity

## Video transcript

Let's say we're going to
shoot some object into the air at an angle. Let's say its speed is s and
the angle at which we shoot it, the angle above the
horizontal is theta. What I want to do in this video
is figure out how far this object is going to travel
as a function of the angle and as a function of the speed, but
we're going to assume that we're given the speed. That that's a bit
of a constant. So if this is the ground right
here, we want to figure out how far this thing is
going to travel. So you can imagine, it's going
to travel in this parabolic path and land at some
point out there. And so if this is at distance 0,
we could call this distance out here distance d. Now whenever you do any problem
like this where you're shooting something off at an
angle, the best first step is to break down that vector. Remember, a vector is something
that has magnitude and direction. The magnitude is s. Maybe feet per second
or miles per hour. And the direction is theta. So if you have s and theta,
you're giving me a vector. And so what you want to do is
you want to break this vector down into its vertical and
horizontal components first and then deal with
them separately. One, to help you figure out how
long you're in the air. And then, the other to
figure out how far you actually travel. So let me make a big version
of the vector right there. Once again, the magnitude
of the vector is s. So you could imagine that the
length of this arrow is s. And this angle right
here is theta. And to break it down into its
horizontal and vertical components, we just set up a
right triangle and just use our basic trig ratios. So let me do that. So this is the ground
right there. I can drop a vertical from the
tip of that arrow to set up a right triangle. And the length of the-- or the
magnitude of the vertical component of our velocity
is going to be this length right here. That is going to be-- you could
imagine, the length of that is going to be our
vertical speed. So this is our vertical speed. Maybe I'll just call that
the speed sub vertical. And then, this right here, the
length of this part of the triangle-- let me do that
in a different color. The length of this part of the
triangle is going to be our horizontal speed, or the
component of this velocity in the horizontal direction. And I use this word velocity
when I specify a speed and a direction. Speed is just the magnitude
of the velocity. So the magnitude of this side
is going to be speed horizontal. And to figure it out,
you literally use our basic trig ratios. So we have a right triangle. This is the hypotenuse. And we could write down
soh cah toa up here. Let me write it down
in yellow. soh cah toa. And this tells us that sine is
opposite over hypotenuse, cosine is adjacent over
hypotenuse and tangent is opposite over adjacent. So let's see what we can do. We're assuming we know
theta, we know s. We want to figure out what the
vertical and the horizontal components are. So what's the vertical component
going to be? Well the vertical component
is opposite this theta. But we know the hypotenuse is
s, so we could use sine because that deals with the
opposite and the hypotenuse. And the sine function tells us
that sine of theta-- actually, let me do this in green since
we're doing all the vertical stuff in green. Sine of theta is going to be
equal to opposite, which is the magnitude of our
vertical velocity. So the opposite side is this
side right here, over our hypotenuse. And our hypotenuse
is the speed s. And so if we want to solve for
our vertical velocity or the vertical component of our
velocity, we multiply both sides of this equation by s. So you get s sine of theta
is equal to the vertical component of our velocity,
s sine of theta. And now for the horizontal
component we do the same thing, but we don't
use sine anymore. This is now adjacent
to the angle. So cosine deals with the
adjacent side and the hypotenuse. So we could say that the cosine
of theta is equal to the adjacent side to the angle,
that is the horizontal speed, over the hypotenuse. The hypotenuse is this length
right here, over s. So if we want to solve for the
horizontal speed or the horizontal component or the
magnitude of the horizontal component, we'd just multiply
both sides times s. And you get s cosine of
theta is equal to the horizontal component. So we now know how fast we are
travelling in this direction, in the horizontal component. We know that that is going
to be s cosine of theta. And we know in the vertical
direction-- let me do that in the vertical direction, the
magnitude is s sine of theta. It is s sine of theta. So now that we've broken up into
the two components, we're ready to figure out how long
we're going to be in the air.

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