# Projectile at an angle | Two-dimensional motion | Physics | Khan Academy

– [Voiceover] So I’ve got a rocket here. And this rocket is going

to launch a projectile, maybe it’s a rock of some kind, with the velocity of

ten meters per second. And the direction of that velocity is going to be be 30 degrees, 30 degrees upwards from the horizontal. Or the angle between the

direction of the launch and horizontal is 30 degrees. And what we want to

figure out in this video is how far does the rock travel? We want to figure out how, how far does it travel? Does it travel? And to simplify this problem, what we’re gonna do is we’re gonna break down

this velocity vector into its vertical and

horizontal components. We’re going to use a vertical component, so let me just draw it visually. So this velocity vector can be broken down into its vertical and its horizontal components. And its horizontal components. So we’re gonna get some

vertical component, some amount of velocity

in the upwards direction, and we can figure, we can use that to figure out how long will this rock stay in the air. Because it doesn’t matter what its horizontal component is. Its vertical component is gonna determine how quickly it decelerates due to gravity and then re-accelerated, and essentially how long it’s going to be the air. And once we figure out

how long it’s in the air, we can multiply it by, we can multiply it by the horizontal component of the velocity, and that will tell us how far it travels. And, once again, the assumption that were making this videos is that air resistance is negligible. Obviously, if there was

significant air resistance, this horizontal velocity

would not stay constant while it’s traveling through the air. But we’re going to assume that it does, that this does not change, that it is negligible. We can assume that were

doing this experiment on the moon if we wanted to have a, if we wanted to view it in purer terms. But let’s solve the problem. So the first that we want to do is we wanna break down

this velocity vector. We want to break down this velocity vector that has a magnitude of

ten meters per second. And has an angle of 30

degrees with the horizontal. We want to break it down it

with x- and y-components, or its horizontal and vertical components. so that’s its horizontal, let me draw a little bit better, that’s its horizontal component, and that its vertical

component looks like this. This is its vertical component. So let’s do the vertical component first. So how do we figure out

the vertical component given that we know the

hypotenuse of this right triangle and we know this angle right over here. And the angle, and the side,

this vertical component, or the length of that vertical component, or the magnitude of it, is opposite the angle. So we want to figure out the opposite. We have to hypotenuse, so once again we write down so-cah, so-ca-toh-ah. Sin is opposite over hypotenuse. So we know that the sin, the sin of 30 degrees, the sin of 30 degrees, is going to be equal to the magnitude of our vertical component. So this is the magnitude of velocity, I’ll say the velocity in the y direction. That’s the vertical direction, y is the upwards direction. Is equal to the magnitude of our velocity of the velocity in the y direction. Divided by the magnitude

of the hypotenuse, or the magnitude of our original vector. Divided by ten meters per second. Ten meters per second. And then, to solve for this quantity right over here, we multiply both sides by 10. And you get 10, sin of 30. 10, sin of 30 degrees. 10 sin of 30 degrees is going to be equal to the magnitude of our, the magnitude of our vertical component. And so what is the sin of 30 degrees? And this, you might have memorized this from your basic trigonometry class. You can get the calculator

out if you want, but sin of 30 degrees is

pretty straightforward. It is 1/2. So sin of 30 degrees, use a calculator if you

don’t remember that, or you remember it now so sin of 30 degrees is 1/2. And so 10 times 1/2 is going to be five. So, and I forgot the units there, so it’s five meters per second. Is equal to the magnitude, is equal to the magnitude

of our vertical component. Let me get that in the right color. It’s equal to the magnitude

of our vertical component. So what does that do? What we’re, this projectile, because

vertical component is five meters per second, it will stay in the air

the same amount of time as anything that has a vertical component of five meters per second. If you threw a rock or

projectile straight up at a velocity five meters per second, that rocket projectile will stay up in the air as long as this one here because they have the

same vertical component. So let’s think about how

long it will stay in the air. Since were dealing with a situation where we’re starting in the ground and we’re also finishing at the same elevation, and were assuming the air

resistance is negligible, we can do a little bit

of a simplification here. Although I’ll do another version where we’re doing the more complicated, but I guess the way that

applies to more situations. We could say, we could say “well what is our “change in velocity here?” So if we think about just

the vertical velocity, our initial velocity, let me write it this way. Our initial velocity, and we’re talking, let me label all of this. So we’re talking only in the vertical. Let me do all the vertical stuff that we wrote in blue. So vertical, were dealing

with the vertical here. So our initial velocity,

in the vertical direction, our initial velocity in

the vertical direction is going to be five meters per second. Is going to be five meters per second. And we’re going to use

a convention, that up, that up is positive and

that down is negative. And now what is going to

be our final velocity? We’re going to be going up and would be decelerated by gravity, We’re gonna be stationary at some point. And then were to start

accelerating back down. And, if we assume that air

resistance is negligible, when we get back to ground level, we will have the same

magnitude of velocity but will be going in

the opposite direction. So our final velocity, remember, we’re just talking about the vertical component right now. We haven’t even thought

about the horizontal. We’re just trying to figure out how long does this thing stay in the air? So its final velocity is

going to be negative five. Negative five meters per second. And this is initial velocity, the final velocity is going

to be looking like that. Same magnitude, just in

the opposite direction. So what’s our change in velocity in the vertical direction? Change in velocity, in

the vertical direction, or in the y-direction, is going to be our final velocity, negative

five meters per second, minus our initial velocity, minus five meters per second, which is equal to negative 10 meters per second. So how do we use this information to figure out how long it’s in the air? Well we know! We know that our vertical, our change our change in our, in

our vertical velocity, is going to be the same thing or it’s equal to our acceleration in the vertical direction times the change in time. Times the amount of time that passes by. What’s our acceleration

in the vertical direction? What’s the acceleration due to gravity, or acceleration that gravity, that the force of gravity

has an object in freefall? and so this, right here, is going to be negative 9.8 meters per second squared. So this quantity over here is negative 10 meters per second, we figured that out, that’s gonna be the change in velocity. Negative 10 meters per second is going to be equal to negative 9.8, negative 9.8 meters per second squared times our change in time. So to figure out the total amount of time that we are the air, we just divide both sides by negative 9.8 meters per second squared. So we get, lets just do that, I wanna do that in the same color. So I do it in, that’s not, well, that close enough. So we get negative 9.8 meters per second squared. Negative 9.8 meters per second squared. That cancels out, and I get my change in time. And I’ll just get the calculator. I have a negative divided by a negative so that’s a positive, which is good, because we want to go in positive time. We assume that the elapsed time is a positive one. And so what we get? If I get my calculator out, I get my calculator out. I have, this is the same thing as positive 10 divided by 9.8. 10, divided by 9.8. Gives me 1.02. I’ll just round to two digits right over there. So that gives me 1.02 seconds So our change in time, so this right over here is 1.02. So our change in time, delta t, I’m using lowercase now but I can make this all lower case. Is equal to 1.02 1.02 seconds. Now how do we use this information to figure out how far this thing travels? Well if we assume that it retains its horizontal component of

its velocity the whole time, we just assume we can this multiply that times our change in time and we’ll get the total displacement in the horizontal direction. So to do that, we need to figure out this horizontal component, which we didn’t do yet. So this is the component of our velocity in the x direction, or

the horizontal direction. Once again, we break out a

little bit of trigonometry. This side is adjacent to the angle, so the adjacent over hypotenuse

is the cosine of the angle. Cosine of an angle is

adjacent over hypotenuse. So we get cosine. Cosine of 30 degrees, I just want to make sure

I color-code it right, cosine of 30 degrees is equal to the adjacent side. Is equal to the adjacent side, which is the magnitude of

our horizontal component, is equal to the adjacent

side over the hypotenuse. Over 10 meters per second. multiply both sides by

10 meters per second, you get the magnitude

of our adjacent side, color transitioning is difficult, the magnitude of our adjacent side is equal to 10 meters per second. Is equal to 10 meters per second. Times the cosine, times the cosine of 30 degrees. And you might not remember the cosine of 30 degrees, you can use a calculator for this. Or you can just, if you do remember it, you know that it’s the square root of three over two. Square root of three over two. So to figure out the actual component, I’ll stop to get a

calculator out if I want, well I don’t have to use it, do it just yet, because I have 10 times the square root of three over two. Which is going to be 10

divided by two is five. So it’s going to be five

times the square root of three meters per second. So if I wanna figure out the entire horizontal displacement, so let’s think about it this way, the horizontal displacement, that’s what we get for it, we’re trying to figure out, the horizontal displacement, a S for displacement, is going to be equal

to the average velocity in the x direction, or

the horizontal direction. And that’s just going to be this five square root of three meters per second because it doesn’t change. So it’s gonna be five, I don’t want to do that same color, is going to be the five square

roots of 3 meters per second times the change in time, times how long it is in the air. And we figure that out! Its 1.02 seconds. Times 1.02 seconds. The seconds cancel out with seconds, and we’ll get that answers in meters, and now we get our calculator

out to figure it out. so we have five time the

square root of three, times 1.02. It gives us 8.83 meters, just to round it. So this is going to be equal to, this is going to be equal to, this is going to be oh, sorry. this is going to be equal to 8.8, is that the number I got? 8.83, 8.83 meters. And we’re done. And the next video, I’m gonna try to, I’ll show you another way

of solving for this delta t. To show you, really, that there’s multiple ways to solve this. It’s a little bit more complicated but it’s also a little bit more powerful if we don’t start and end

at the same elevation.

NICE VID SAL

By the way, an easy way to fin Range is:

Range=sin(2(theta))*(velocity)^2/(gravitational acceleration)

thank you so much for making these videos

Your a lifesaver.

@khanacademy

hi sal ,

so i have a question about the vertical velocity , correct me if i was wrong so to get the normal average velocity we do it like this ''delta''V =(Vi+Vf)/2 , but here when you calculated the vertical velocity which is also ''delta''V you calculated it like this ''delta''V = (Vf-Vi) and you used the same formula for the average velocity which is: a=''delta''V/''delta''t so whtat's the difference between those two velocities ?

How do i find the distance and initial velocity if only the angle and height is given?

TI-85!

hi can someone explain why the magnitude of the final velocity is equal to the initial velocity?

very good tutorial…you are doing a very good work!!

@rdcruick watch the previous videos of projectile motion added by sal sir in the playlist physics it willl explain well

@rdcruick /watch?v=T0zpF_j7Mvo

really ur helping me a lot…..right now..esp.. wen i dont get concepts in scool…

my teacher doesnt teach… i thought i was retarded until i saw this vid=/

I feel like if physicists can calculate gravity, they should be able to settle on some sort of constant for air resistance. Most of these problems are not valid in the real world.

The thing is air resistance depend of many things like the shape of the objet for example, so it can't be a constant. Plus when you fall you gain speed and the faster you are the more air resistance you get, so you'll have to use a differential equation. They are ways to calculate it but it's kind of complex and take a lot of paradigms so physic problems in school usually don't bother with it.

Air resistance is dependent on many factors including the temperature, speed and the object's drag coefficient.

may GOd bless you

I a bit confused about the triangle in the video. If the hypotenuse it 10 and it is a right triangle then shouldn't the sides fulfill the 3:4:5 ratio? If the angle is 30deg and it is a right triangle then shouldn't it fulfill the 30,60,90 right triangle pattern? It seems like the two patterns are mutually exclusive…I get a different (incorrect?) answer if I solve the problem using the 3:4:5 ratio. can anyone help me reconcile this?

man you're the reason why i have a 90 in physics. thank you

I did not expect the giant calculator lol

You're*

What program do you use to make these videos? i've been really curious…

i could learn more and faster if i just followed my curriculum with his videos…id actually learn something

how would you calculate initial velocity if you know the launch angle and the time that its in the air?

I like that candy too!

Can anyone help me I am still lost!

let me summarize this. (khan does not use the most direct way of solving this, but its kind of cool nonetheless).

We are given that the initial velocity of a particle has a magnitude of 10 meters/second, and points at an angle of 30 degrees with the horizontal to the right (so this is a vector in 2 dimensions, and we don't need a third dimension). We want to find the total distance this particle covers under only the influence of gravity (no air resistance involved).

First thing we want to do is break down the velocity vector

vinto horizontal and vertical components. That way we can recast a 2-d problem into two 1-d problems, since we know our motion equations for 1-d motion.So again, we are given that ||

v||= 10, and theta = 30 degrees. So we can use our knowledge of trigonometry.v_x = ||

v|| cos thetav_y = ||

v|| sin thetav_x = 10 cos 30 = 8.66 m/s

v_y = 10 sin 30 = 5 m/s

Now lets look at the vertical and horizontal motion separately.

We know that

delta v_y = <a_y> delta t, where <a_y> means average acceleration in the vertical direction (not instantaneous acceleration). But we know that for vertical motion in free fall, the acceleration due to gravity is a constant -9.8 m/s^2.

Therefore we can substitute the constant 'a' of -9.8 for <a_y>.

And we also know that v_yi = 5 m/s and v_yf = -5 m/s by symmetry (sal explains it in the video).

v_yf – v_yi = a* delta t

(-5) – 5 = -9.8* delta t

So delta t = 10/9.8 = 1.02 seconds, which is how long the particle is in the air.

Now we also know that

delta s_x = <v_x>*delta t.

But here v_x is constant so <v_x> = v_x , since gravity has no effect on an objects horizontal motion, only vertical. So plugging in we have

delta s_x = 8.66 * delta t

delta s_x = 8.66 * 1.02 = 8.8 meters.

Thank god for this video; everything makes so much more sense now 😮

For me it has vivem a slighty diferent answer. But I multiplicated five by the square root of three. What is going onde here?

How will you solve the problem if you dont have the angle?!?!?!?!?????!!!!!

can someone explain to me why he chose to use -9.81 (at 7:53) and not positive 9.81? 🙁

Saved my life

I wish I had found it before taking my exams, educational nonetheless! 🙂 Great video. Looking forward to learning a lot!

one question, wont g be negative when y direction moves up and then positive when it moves down? you cant take it to be -ve for both the sides!!

Man I only have the angle degrees but no velocity…. Idk how to start it.

Sal, you are so amazing! Its appreciable how one person can explain so many subjects in detail!! Videos by Khan Academy have helped me in understanding a lot of stuff and will always help me in the future! Thank You!!

Thank you so much.

How in the world does he get those little formulas? I know Vf=vi+at but where is he getting those delta from?

kahn academy's success is exemplary of public school's failure.

中国人民感谢可汗学院

absolute bull

wouldnt the intial velocity be 0 m/s???

color transitioning is difficult. hahahaahah

Sal's intelligence gets me a hard on

hey won't you take final velocity as 0

physics is hard

you saved my life thanks

Why can we use 10m/s as a length for the hypotenuse when its just speed in a given direction? I know its a vector quantity but it doesnt quite make sense to me..

can anyone tell me why sin30 = 1/2 in this problem?

how did you find the time(delta t)

why not to know the hang time by that rule y=(viy)(t)

~~1/2 (g)(t)power 2 like for the first example we have 10 and 30○ :~~0=10sin(30)(x)-1/2 (9.8)(x)power 2 and shift solve on the calculator =1.0204 as we know everything expect the t that we already trying to get g=is always 9.8This really makes me understand how science and maths suddenly become relevant to each other.

if you want the range just use this formula R=vo^2([email protected])g .

@= means the theta angle or whatever angle name

it it the fastest way to solve the questing

just use the formula

Where does the equation;

Vy = ay.T come from???

That's not even an equation

Wow I liked that calculator, can someone tell me where to get it?????

So good.

my 4 year old bought me here. He's dead set about trajectories… when racing hot wheel in midair… I think i need to get him a physics & math tutor. What 4 year old does that?

What if you are only given the total displacement and the angle?

All 35 dislikes are by people who make similar videos but charge for it 😂

All their videos are helpful but damn i couldn't understand this

i love you

So how would I calculate the angle of initial projectile and angle of impact on a parabolic trajectory? If I were given a ballistics chart based on known drag coefficient specs? For example trajectory leaves at 290 feet traveling a distance of 2000 feet hitting an object at 30 feet from the ground…

Sorry, but you are over complicating things here. IN this example we can use the vertical component to calculate the time for the vertical flight. Because the ball lands on the same level we use SUVAT for the vertical component and that gets you the answer.

At about 7:10, he writes this formula:

Delta-Vy = acceleration-y times delta-t.

From where does he get this formula?

The closest thing I can find is one of the four kinematic equations, V=V0+at. However, initial velocity in this example is not zero. It is 5 m/s. Therefore this is not the same equation. It's also not the displacement over velocity formula, or the acceleration equals average velocity over time formula. As far as I can tell this formula came from nowhere. Where did it come from? What is it?

After a million hours of one dimensional motion, two is easy lol

how do you write so good with your mouse?

when i try, i write HRLO instead of HELLO

Range=V^2.sin(2Q)/g = use this equation to find horizontal displacement

you guys are my hero

Is there a video somewhere on describing the flight of a rocket launched at an angle with constant thrust and linearly decreasing mass?

time in the air is time up + time down i think you are missing something

Why the fuk professors can't teach like that?

This makes no sense to me rn. Ugh why did i take honors chem and honors physics at the same time

You could use Pythagoras theorem to get the Vx.

Legend

cos 30 is √3/2, not √3 🙂

how the hell speed becomes a hypotnuse..i mean hypotenuse is just for lengths

I didn't understood how initial and final velocity are same, if some one knows please explain

OR you can use the Range Equation. its a whole lot faster and its applicable due to y's displacement being 0.

R= (Vo^2*sin2theta)/g R=((10^2)(sin(60)))/9.8 = 8.83m

wait if there is acceleration due to gravity in the "y" then how is the initial and final velocity the same magnitude?

1.25 speed

i still wonder how he films these videos

Nyc

Fck this physics HAHAHAA

Sal Khan is the greatest gift to us mortals. We are unworthy of his teaching effectiveness.

sal calculated the time it takes to go up to max height amounting to about 1 sec.

however, the time taken to reach the ground is twice the time taken to reach max height

this means time of flight is 2 seconds

so the distance should be twice the value calculated i.e 17.6 m

please explain if I'm right or where i've gone wrong

Game of Thrones brought me here.

WOW! THAT WAS AMAZING!!

nice

Why is the final velocity -5m/s? Wouldn't it be 0m/s at it's final velocity once it hit's the ground?

How is the initial velocity and final velocity in vertical direction same in magnitude?

Is it because the initial and final position is the same?

Meanwhile:

Katy Perry:

Is sCiEnCE RElaTeD tO MaTh !!!???4:07 i got -0.9880316241