Let's do a slightly more complicated 2D projectile motion problem now

So in this situation I am going to launch the projectile off of a platform

And then it is going to land on another platform

It's coming out of the cannon. Let me do it this way just to make it 100% clear

So this angle right over here is 53 degrees

And we are going to come out the muzzle of the cannon with a velocity of 90 m/s

Just to give ourselves a sense of how high it's being launched from

From the muzzle of the cannon down to here, so this height right over here is 25 meters

and let's say that this height right over here is 9 m

And so we are essentially launching this from a height of 25 meters

I know in the last video even though I drew the cannon like this

we assumed that it was being launched it from a altitude of 0

and then landing back at an altitude of 0

Here we are assuming we're launching it from an altitude of 25 meters

That's when it's leaving the muzzle

and it's gonna start decelerating at least in the vertical direction

as soon as it leaves the muzzle

and then we're assuming it's not gonna land back at the same altitude

It's gonna land at a different altitude

So how do we think about this problem?

So the first thing you always wanna do

is divide your velocity vector into its horizontal and vertical components

You use the vertical component to figure out how long it will stay in the air

And then you use the horizontal component to figure out

given how long it's in the air, how far does it travel?

And once again, we're gonna assume the air resistance is negligible

So just based on what we did in the last video

I'll go through all of these steps in this one as well

If we draw our vector

the length here is going to be 90

The angle over here--this is our velocity vector--the angle over here between

And let me draw the horizontal component. It would look like this

And the vertical component would look like this

And so the vertical component of the vector, what would be the side right over here?

Well, this is the opposite side

We know from basic trigonometry sine of an angle is opposite over the hypotenuse

is equal to the magnitude of the vertical velocity--

I write that subscript Y, for the Y direction. That's the vertical direction

Over the length of the hypotenuse, over the magnitude of our original vector

Or if we multiply both sides by 90

we get that the magnitude of that side

Now if we want to do the horizontal component

the horizontal side is adjacent to this

Soh-cah-toa, cos is adjacent over hypotenuse

So the horizontal component of our velocity, I'll say x direction

over the hypotenuse, over 90

Cosine is adjacent over hypotenuse. Adjacent that's this length over 90

Multiply both sides by 90, you get that the horizontal component

Now how do we figure out how long the thing takes in the air?

We'll use the vertical component for that

Especially since we're dealing with different levels, we can't use that

more basic reasoning that, "Hey, whatever velocity we start off at

it's gonna be the same magnitude of velocity but the opposite direction

because we're not going to the same elevation

What we could do is we can use the formula that we derived in the previous video

that the displacement--let me just copy and paste this a little bit lower

Copy it and paste it. I'll get it right over here

So we can use this

We know that the displacement is equal to the initial velocity

and we're dealing with the vertical direction right here, times the change in time

plus the acceleration times the change in time squared divided by two

So how do we use this to figure out how long we're in the air?

So what is the displacement if we're starting at 25 m high and we're going to 9 m high?

So over the course, while this thing is traveling it will be displaced downwards 16 m

Or another way to think about it is

our displacement in the vertical direction is going to be equal to -16 m

Right? Because 25-9 is 16

And so we can put that into the formula we derive in the previous video. We get

-16 -- I won't write the units here for the sake of space and simplicity--

is equal to the initial velocity--we're dealing with just the vertical dimension here

Remember, it's negative because our displacement is going to be downwards

We're losing altitude

Times our change in time

[plus] the acceleration due to the force of gravity for objects in free fall

It's -9.8 m/s squared

but we're dividing that by two. So we have -4.9 m/s^2

times delta t squared, times our change in time squared

So how do we solve something like this? You can't just factor out a t and solve it

You might recognize that this is a quadratic equation right over here

The way you solve quadratic equations is

you get everything on one side of this equation and then

you either factor it out, or more likely in the situation, you will use the quadratic formula

which we proved in other videos, hopefully, giving you the intuition for it

to actually solve for the times where your displacement in the vertical direction is -16 m

I will get 2 solutions here, and one solution will be a negative change in time

There some time in the past, you're also at -16 m

That's nonsensical for this problem

So we want to take the positive value here

So let's put all of this on one side of the equation

Let's add 16 to both sides

On the left-hand side you get a zero

Zero is equal to--I'll write it in the traditional way that we're used to seeing

I'll write the highest degree term first. So -4.9 times delta t squared

and then +16

All of this is equal to zero

And this once against is just the quadratic equation. We can find its roots

and the roots will be in terms of delta t. We can solve for delta t using the quadratic formula

So we get delta t--

if this is very unfamiliar to you, review the videos on Khan Academy algebra playlist

on the quadratic formula. If you don't know where it came from, it also proved it for you

So it's equal to negative b--b is this right here, the coefficient on the delta t

I'll write the quadratic formula for those of you who can't remember it

So I'm going to solve Ax^2+Bx+C=0

The roots over here is going to be -B

These are going to be the X values that satisfy this equation up here

So that's what I'm doing over here. This is the B value, negative B plus or minus--

It turns out we only care about the plus one, because that's gonna give us the positive value

B squared, so it's this quantity squared

-4 times A which is -4.9, times C which is 16

This radical all the way over here. All that over 2A

A is -4.9, so 2A is -9.8

So now we can get the calculator out to figure out our change in time

I'm just gonna focus on the positive version of it

I'll leave it up to you to find the negative version and see if

that'll give you a negative value for change in time

And that's nonsensical. So we only care about the positive change in time where we get to

a placement of -16 meters

Let's get the calculator out

So we get--let me do this carefully. We have

these two negatives cancel out. So it's plus 4 times positive 4.9 times 16

and that closes off our entire radical

And so this will give me the numerator up here

And I want to divide that by -9.8

Oh, I just realized that I made a mistake

I said that the positive version would give you the positive time

but now we realize that is wrong because when I took the positive version up here

I get a positive 2.14 for the numerator, but then we divide it by -9.8

we'll get a negative value. So that's not gonna be the time we care about

So we care about the time where this is a negative value

So let me reenter that. Let me do the negative value

Let me go back a little bit

And then let me replace this with a -

So we look at the negative value, because I want the positive time

And so now my numerator here is a negative value. So this is actual what we care about

The numerator is a negative value, you divide it by -9.8

and you get--I'll just round--14.89 seconds

So delta t the positive version is equal to 14.89 second

So my initial comment about wanting to using the positive version was wrong

because we have a negative denominator, so you want the numerator to be

negative, and only the numerator is negative, will the whole expression be positive

So we've got this positive time of 14.89 second

Let me solve for the horizontal displacement, although this is running long

So the amount of time that we're in the air is 14.89 seconds

So if I were to ask you the horizontal displacement

it's going to be the amount of time we're in the air

times your constant horizontal velocity

We already figured it out, our constant horizontal velocity

So if you want to figure out how far along the x-axis we get displaced

we just take this time times--that just means our previous answer--

and that gives us 806 m. So this displacement right over here is 806 m