# Projectile motion (part 1) | One-dimensional motion | Physics | Khan Academy

Welcome back. I’m not going to do a bunch of

projectile motion problems, and this is because I think

you learn more just seeing someone do it, and thinking

out loud, than all the formulas. I have a strange notion that I

might have done more harm than good by confusing you with a lot

of what I did in the last couple of videos, so hopefully

I can undo any damage if I have done any, or even better–

hopefully, you did learn from those, and we’ll

just add to the learning. Let’s start with a

general problem. Let’s say that I’m at the top

of a cliff, and I jump– instead of throwing something,

I just jump off the cliff. We won’t worry about my motion

from side to side, but just assume that I go

straight down. We could even think that someone

just dropped me off of the top of the cliff. I know these are getting kind

of morbid, but let’s just assume that nothing

bad happens to me. Let’s say that at the top

of the cliff, my initial velocity– velocity initial– is

going to be 0, because I’m stationary before the person

drops me or before I jump. At the bottom of the cliff

my velocity is 100 meters per second. My question is, what is the

height of this cliff? I think this is a good time

to actually introduce the direction notion of

velocity, to show you this scalar quantity. Let’s assume up is positive,

and down is negative. My velocity is actually 100

meters per second down– I could have assumed

the opposite. The final velocity is 100 meters

per second down, and since we’re saying that down

is negative, and gravity is always pulling you down, we’re

going to say that our acceleration is equal to

gravity, which is equal to minus 10 meters per

second squared. I just wrote that ahead of

times, because when we’re dealing with anything of

throwing or jumping or anything on this planet, we

could just use this constant– the actual number is 9.81, but

I want to be able to do this without a calculator, so I’ll

just stick with minus 10 meters per second squared. It’s pulling me down, so that’s

why the minus is there. My question is: I know my

initial velocity, I know my final velocity, right before I

hit the ground or right when I hit the ground, what’s

the distance? In this circumstance, what

does distance represent? Distance would be the height of

the cliff, and so how do we figure this out? What’s the only formula that

we know for distance, or actually the change in distance,

but in this case, it’s the same thing. Change in distance is equal

to the average velocity. When you learned this in middle

school, or probably even elementary school, you

didn’t say average velocity, because you always assumed

velocity was constant– the average and the instantaneous

velocity was kind of the same thing. Now, since the velocity is

changing, we’re going to say the average velocity. So, the change in distance is

equal to the average velocity times time. This should be intuitive

to you at this point. Velocity really is just distance

divided by time, or actually, change in distance

divided by times change in time– or, change in

distance divided by change in times is velocity. Let me actually change this

to change in time. Since we always assume– or we

normally assume– that we start at distance is equal to 0,

and we assume that start at time is equal to 0, we can write

distance is equal to velocity average times time. Maybe later on we’ll do

situations where we’re not starting at time 0, or distance

0, and in that case, we will have to be a little more

formal and say change in distance is equal to average

velocity the change in time. This is a formula we know,

and let’s see what we can figure out. Can we figure out the

average velocity? The average velocity is just

the average of the initial velocity and the

final velocity. The average velocity is just

equal to the average of these two numbers: so, minus 100 plus

0 over 2– and I’m just averaging the numbers– equals

minus 50 meters per second. We were able to figure

that out, so can we figure out time? We know also that velocity,

or let’s say the change in velocity, is equal to the

final velocity minus the initial velocity. This is nothing fancy– you

don’t have to memorize this. This hopefully is intuitive to

you, that the change is just the final velocity minus the

initial velocity, and that that equals acceleration

times time. So what’s the change in velocity

in this situation? Final velocity is minus 100

meters per second, and then the initial velocity is 0, so

the change in velocity is equal to minus 100 meters

per second. I’m kind of jumping in and out

of the units, but I think you get what I’m doing. That equals acceleration

times time– what’s the acceleration? It’s minus 10 meters per second

squared, because I’m going straight down– minus 10

meters per second squared times time. This is a pretty straightforward

equation. Let’s divide both sides by the

acceleration, by the minus 10 meters per second squared, and

you’ll get time is equal to– the negatives cancel out, as

they should, because negative time is difficult, we’re

assuming positive time, and it’s good we got a positive

time answer– but the negatives cancel out

and we get time is equal to 10 seconds. There we have it: we figured out

time, we figured out the average velocity, and so now we

can figure out the height of the cliff. The distance is equal to the

average velocity minus 50 meters per second times

10 seconds. The distance– this is going to

be an interesting notion to you– the distance it’s going

to be minus 500 meters. This might not make a lot of

sense to you– what does minus 500 meters mean? This is actually right, because

this formula is actually the change

in distance. We said if we did it formally,

it would be the change in distance. So if we have a cliff– let me

change colors with it– and if we assume that we start at this

point right here, and this distance is equal to 0,

then the ground, if this cliff is 500 hundred meters high, your

final distance– this is the initial distance– your

final distance df is actually going to be at minus

500 hundred meters. We could have done it the other

way around: we could have said this is plus 500

meters, and then this is 0, but all that matters is really

the change in distance. We’re saying from the top of the

cliff to the ground, the change in distance is

minus 500 meters. And minus, based on our

convention, we said minus is down, so the change is 500

meters down, and that’s height of the cliff. That’s pretty interesting. If you go to a 500 meter cliff–

500 is about 1,500 feet– so that’s roughly the

size of maybe a very tall skyscraper, like the

World Trade Center or the Sears Tower. If you jump off of something

like that, assuming no air resistance, which is a big

assumption, or if you were to drop a penny– because a penny

has very little air resistance– if you were to drop

a penny off of the top of Sears Tower or a building like

that, at the bottom it will be going 100 meters per second. That’s extremely fast, and

that’s why you shouldn’t be doing it, because that is fast

enough to kill somebody, and I don’t want to give you any bad

ideas if you’re a bad person. It’s just interesting that

physics allows you to solve these types of problems. In the next presentation, I’m

just going to keep doing problems, and hopefully you’ll

realize that everything really just boils down to average

velocity– change in velocity is acceleration times time,

and change in distance is equal to change in time times

average velocity, which we all did just now. I’ll see you in the

next presentation.

for convenience it is taken as 10ms^2 just as speed of light is 299,792 m/s but it`s mostly taken as 3 X 10^6 or 3000000 m/s

I know that you think that the penny would kill you how ever it wouldn't due to updrafts against the building. I just like to correct hypothetical situations :P.

For most problems, it's advisable to use average value of 9.8 m/s^2. It's average since g varies over the surface of the earth for various reasons. One can use g as to 10 m/s^2, as in this case, for doing a rough order of magnitude estimation. Hope it helps.

0_0 this is physics, EVERYTHING I KNOW IS A LIE!

like a sir ~~(mustache)

This guy knows EVERYTHING! Scary…

Physics is math.

i know i was being sarcastic

or math-based physics… o_O

distance is scalar quantity why you don't put d as displacement?(because displacement is vector have direction and magnitude)

When you learn this is middle school, elementary school even… dude I didn't even hear any of these terms until my 3rd year of high school..

It feels ridiculous to be paying so much for university classes when I learn much of the subjects from Sal. If anything he is much more deserving of my money.

I agree that d could (should) be displacement… I was thinking that in my head the whole time, so I'm glad someone else was/it was right… just vertical displacement, right?

I have a question:

How does air resistance play a role in solving the problem?

Why couldn't you just use v^s=u^2+2as?

Sal dont jump off cliffs man youtube kinda needs you

How is it that you decide whether to use average velocity or change in velocity?

It doesn't. He is trying to simplify it as much as he can as a teacher in a classroom would simplify it. He is teaching you just the basics. Air resistance is where it starts getting more complicated so that is taught later.

i want sex with u 🙂

i thought it was -9.8 m/s^2?

"these are getting kid of morbid" ….lol

he just rounds to 10 because it's more convenient to do/teach, but when actually doing it, it should be -9.8.

why is the distance negative? isnt distance a scalar quantity not a vector so it should just be 500m?

Mythbusters proved that a penny dropped at 100 m/s cannot even puncture the skin…

That's because of the momentum 🙂

No sal don't jump!!!!!!!!!

nice

very nice lecture

Man make your drawing more clear and put the Delta, its sometime confusing

uh universal gravitation is actually -9.81m/s^2

They often make it simple by using 10 instead of the actual number

And actulally its 9.80665 so you are not entirely correct yourself seen as they have just rounded it

Because the way he used the formulas implied that he had superimposed a coordinate system over the cliff, where the edge of the cliff was the origin, and anything below that would therefore yield negative y values…

Isn't Vavg=delta x /delta t?

@jejealam. That's true, because Delta X represents the change of displacement.

i love it 🙂

I see the light!

why didn't you simply use V^2 – U^2 = 2as,,,,,,ehh.,! srry if i sound rude.,! 😀

V-final velocity,

U-intial velocity,

a-acceleration,

s-distance,

the acceleration due to gravity should be positive….

Did yoy major in like All subjects or what… You're amazing! 🙂 thank you

the "-" stands for going down

Can't you use the timeless equation

LOLOLOLOLOLOL

(05:09)

why u adding = a . t in changing of velocity, im confuse cuz it is not even in the formula it self, i know what you trying to find , but i just don't get how u put that there ?

I hope you mean "d" as in displacement. There is no way that distance can be negative.

If he is falling down a cliff, I think his final velocity is also zero 🙂

I did it with kinetic and height energy, and found that he weighs ( at least in this equation ) 100kg

Because I inserted that and I got the exact same answer

I get it now. thank you so much.

My teacher teach me to use a=g=9.81.. its it same??

thanks it helps me to physics thumps up 🙂 👍

@aiman fitri

ais a general notation for "acceleration", whereasgis specifically used for the acceleration ofgravity.You can use

ainstead ofg, as long as it is clear that you mean the acceleration of gravity.Alternatively, you can write

awith the subscriptg.#aiman fitri it is -9.8m/s^2. He just rounded it to -10m/s^2

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Sal, I just wanted to comment on what you said about dropping the penny from the top of a tall building. You mentioned that it would go fast enough to kill or hurt someone. However this is not true because the penny does not have enough mass to gain a velocity that would be harmful, even if it was dropped from a thousand meters. Let me know what you think. I am sure you were just giving an example but I just wanted to clarify it for you..

Can't you just use the equation Vf^2=Vi^2+2ad???

Did anyone else cringe at the rounding of g to -10ms^-2? xD

What is the difference between changing velocity and average velocity? Why you can not use average velocity to calculate time?

isnt when we go towards gravity the value is positive 10m/s^2?? why did he take negative??

"When you learned about average velocity in elementary school"

WHAT?? WHAT KIND OF ELEMENTARY SCHOOL DID YOU GO TO???

whos watching this for ms.cole?? lmaaoo

who else here about to stay up all night!!!!

wouldnt it be displacemnt instead of distance

Acceleration due to gravity is not 10. It is 9.8

Shouldn't it be displacement?

Good video but please use correct terms and signs. You may confuse many students. Like say displacement and not distance. And use vector signs.

I don't wanna give you a bad idea if you are a bad person.

After 10 years, this is still the best video I've seen on this topic! 😀

Thank you sal khan! I'm in fifth grade and this stuff makes sense!

Sal's voice in 2007 videos is just so calm and idk alluring, I often find myself getting distracted. Which really shouldn't happen, because Khan Acad's education is gold too 👌

Khan Academy I would hug you as tightly as I can because you've helped me a lot with this video Thank You So Much! 🙏😊

Isn't acceleration downwards +10m/s(2) because If it's going upwards it should then be -10m/s(2) because it's gravity, gravity is always downwards

This is not what want… I need the courses of India and nepal Cbse and hseb board

Physics exam tomorrow, all-nighter let's gooo

I hope you put the standard formulas

Never heared about this subject. Watched this video and already understand it😂😂😂

How does final velocity – initial velocity= acceleration . time

You, Sir, are amazing. That is all.

sal seems sad in this video.

Terminal velocity. Can't kill anyone with a penny. Need something with more mass, like a piano or a nuke maybe.

you could jus use 2as=v^2-u^2

I love learning in 240p quality.

i love Khan academy's videos but this on isn't up to the mark….

Who else feels the same???

Sir, i have doubt.DISTANCE IS A SCALAR QUANTITY BUT IN AN EX THE S WAS -500m. can scalars be negativ?

First ap physics test tomorrow, hopefully sal can save my grade lol.

shouldn't he be using one of the kinematic formulas? I know that at least in Canada that projectile motion is taught after kinematic equations so wouldn't it be easier to use one of them?

What if ur given mass?

Really amazing

We should experiment falling of from a cliff

ugh the video quality…

Hey, thanks bro I am now a topper in my class

How many teachers are in Khan academy Sal khan and why do you only upload on this channel