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### Course: High school physics > Unit 3

Lesson 5: Projectiles launched at an angle# Optimal angle for a projectile part 1: Components of initial velocity

You want a projectile to fly as far as possible, at which angle should you launch it? We'll start with formulas for the initial velocity. Created by Sal Khan.

## Want to join the conversation?

- What do you do if the problem given does not have an angle? What equation do you use to find the angle?(11 votes)
- I think you can use reverse trigonometry, if it gives you two sides you can use Pythagorean because it'll be a right triangle, and you can plug in the side lengths for reverse trigonometry.(6 votes)

- What do scos and ssin mean?(9 votes)
- s is the velocity of the thrown object

scos is the horirontal component of the velocity while ssin is the vertical component of the velocity(3 votes)

- i noticed that you use the theta symbol and i would like to know what it means.(4 votes)
- It's just that: a symbol. Similar to x or y. It is Greek. It is commonly used to represent an angle.(12 votes)

- How would the problem change if you are given an initial or final height of something other than 0?(7 votes)
- how do you derive the maximum height reached by the object?(4 votes)
- I'm assuming you want the answer, and not a mathematical proof.

For just a quick review, the three most important equations in projectile motion are:

Δx = vt + (1/2)at^2

vfinal = vinitial+ at

(vfinal)^2 = (vinitial)^2 + 2aΔx

The maximum height of a projectile can be found from the formula (v)^2 = 2aΔy, where v is the initial vertical velocity of the projectile and a is the acceleration (most often 9.8 m/s/s, or "little g"). The formula can be rearranged to find the vertical displacement (maximum height):

Δy = v/2a(3 votes)

- how do you write maximum height?(3 votes)
- Why are we using speed instead of velocity again?(3 votes)
- why wouldn't it be velocity if it was a speed with a direction?(2 votes)
- When teacher want talk only about magnitude of velocity he use word "speed". Additional information, If your solving 1 dimensional vector problems including velocity you can keep the arrows at the top off they are not needed your sign take care of direction .(2 votes)

- can you find an angle if only given the the total time and total distance?(2 votes)
- Well first break up the initial velocity into components:

vx = vi cos θ (velocity in x direction)

Solve for vi:

vi = vx/cos θ

We know distance and time, so vx = d/t which is a value.

So vi = (d/t) / cos θ

vy = vi sin θ (initial velocity in y direction)

Substitute vi:

vy = [(d/t) / cos θ] sin θ

a = -9.8 m/s^2 is assumed

So we have initial velocity in y direction, acceleration, and time.

We can use the kinematic equation: vf = vy + a*t. If we substitute in what we know there are still 2 unknowns, vf and θ. What to do? We need another equation...

I know! We should halve the time. This is when we know vf = 0 m/s as the object is at its maximum height:

How about now?

0 = [ (d/t) / cos θ ] sin θ + a * (t/2)

The only unknown is θ. Did we do it?? Test it on a problem that has all these variables!!(2 votes)

- A projectile is thrown with an initial velocity which has horizontal component of 4 m/s. What will be its horizontal speed after 3 seconds?(0 votes)
- The horizontal velocity will not change since the acceleration from gravity only acts on the vertical component of the velocity.(6 votes)

## 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.