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### Course: Class 11 Physics (India)>Unit 8

Lesson 8: Optimal angle for a projectile

# 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?
• 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.
• What do scos and ssin mean?
• 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
• i noticed that you use the theta symbol and i would like to know what it means.
• It's just that: a symbol. Similar to x or y. It is Greek. It is commonly used to represent an angle.
• How would the problem change if you are given an initial or final height of something other than 0?
• how do you derive the maximum height reached by the object?
• 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
• how do you write maximum height?
• Why are we using speed instead of velocity again?
• why wouldn't it be velocity if it was a speed with a direction?
• 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 .
• can you find an angle if only given the the total time and total distance?
• 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:
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!!
• 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?
• The horizontal velocity will not change since the acceleration from gravity only acts on the vertical component of the 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.