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## Class 11 Physics (India)

### Unit 8: Lesson 8

Optimal angle for a projectile- Optimal angle for a projectile part 1: Components of initial velocity
- Optimal angle for a projectile part 2: Hangtime
- Optimal angle for a projectile part 3: Horizontal distance as a function of angle (and speed)
- Optimal angle for a projectile part 4: Finding the optimal angle and distance with a bit of calculus

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# 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's the diffrence between speed and velocity?!(14 votes)
- also, if you are talking about average speed and average velocity, average speed can never be zero, whereas average velocity can be zero. This is because Average Speed = total distance traveled/total time taken and average velocity = displacement/time taken. Displacement can be zero because it is the shortest distance between two point and in case the moving object returns to it's initial position then displacement is zero. Hope that helps and adds to your question.(13 votes)

- What do you do if the problem given does not have an angle? What equation do you use to find the angle?(8 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.(4 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.(3 votes)
- It's just that: a symbol. Similar to x or y. It is Greek. It is commonly used to represent an angle.(8 votes)

- How would the problem change if you are given an initial or final height of something other than 0?(6 votes)
- how do you derive the maximum height reached by the object?(3 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(2 votes)

- how do you write maximum height?(2 votes)
- Would the distance be any different if there was any wind, or would it change if the power of what was being launched was changed? And just a quick thing as well, the said "projectile" was not specified if it was or wasn't round, cause if it was round wouldn't it roll making the distance even farther than it was when it first hit the ground? Unless it was say a box that was launched, I do belive I have just confuesed myself in keeping this straight, I don't quite understand stuff like this to well, so answers would be nice. :D(2 votes)
- We only calculate the distance the moment it hits the ground.(2 votes)

- at1:04Sal says that the direction is theta, but the diagram depicts theta as being the angle at which the projectile is thrown, Which is the right one?(1 vote)
- They're the same thing! If an angle relative to the horizontal axis (x-axis) is given along with a line (in this case vector s), then we know the direction of that line. If you think about it, these two bits of information (angle and magnitude) can tell us a lot about an object's trajectory. Theta is just a term for an angle for which we don't have a precise figure.(2 votes)

- Hi everyone! I was wondering - the shape of the trajectory is parabolic, so why do we make calculations using a triangle with straight edges (because the trajectory isn't a straight line). This question especially applies to where the object is launched directly from the ground. Do we just imagine a triangle a bit behind where the trajectory starts?(1 vote)
- the rays forming that triangle are tangent to the parabola, so it's an estimate of the angle measure at the beginning of the trajectory.(2 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.