# 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
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
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.

## 100 thoughts on “Projectile motion (part 1) | One-dimensional motion | Physics | Khan Academy”

1. jensons88 says:

well explain. you can watch a program that calculate projectile motion before the object is released. it's on my channel. you can also download the program from softpedia.

2. cuntraband says:

Man, this really helped. I've been so lost in my physics class. These videos are saving my grade c:

3. Mannan Rehbari says:

i dont think any single video of Sal deserves a dislike… do you?

4. Smidz018 says:

this is motion basics not projectile motion..

5. ambarigreyson3000 says:

bro u cant drop an apple????

6. Syed Sabeel Hasan Shah says:

@Xoneize average velocity can be both defined by average distance travelled in time or (vf+vi)/2. No difference.

7. Ved Pyakurel says:

Please dont jump.. who is gonna make videos if you jump..??

8. Julz333666 says:

@gotwhiteboy09 fuck off idiot.

9. Akshay Shukla says:

then i suppose the formula
v^2-u^2=2as

v: final velocity
u: initial velocity
a: accerlation
s: distance

i got 500 ansert straight?

10. Akshay Shukla says:

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

11. F JOM says:

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.

12. Race Bannon says:

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.

13. TechN9ician2159 says:

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

14. Lohanne Moretz Vandeviere says:

like a sir ~~(mustache)

15. Ezra P. Pierre says:

This guy knows EVERYTHING! Scary…

16. eatcarpet says:

Physics is math.

17. TechN9ician2159 says:

i know i was being sarcastic

18. Jake B says:

or math-based physics… o_O

19. Interception master says:

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

20. Theory says:

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..

21. Sam Champine says:

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.

22. Monica J says:

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?

23. TheRautfamily says:

I have a question:
How does air resistance play a role in solving the problem?

24. TEN YOLO says:

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

25. imthetitsDD says:

Sal dont jump off cliffs man youtube kinda needs you

26. Beza Abegaz says:

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

27. Zac4548 says:

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.

28. Meghna Goswami says:

i want sex with u 🙂

29. lolidc says:

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

30. maksmiles says:

"these are getting kid of morbid" ….lol

31. bobthespin says:

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

32. TheGeniusism says:

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

33. Chris Dawson says:

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

34. quangluu96 says:

That's because of the momentum 🙂

35. M A says:

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

36. mukesh kumar says:

nice

37. mukesh kumar says:

very nice lecture

38. Kashuf Karim says:

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

39. Corey Li says:

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

40. Olly Metcalfe says:

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

41. Olly Metcalfe says:

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

42. Juxtaroberto says:

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…

43. Mocapie says:

Isn't Vavg=delta x /delta t?

44. Chukwuma Aneke says:

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

45. Aishi Aratrika says:

i love it 🙂

46. Lord McSwain says:

I see the light!

47. siddhant pandey says:

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,

48. Abel Fulgencio says:

the acceleration due to gravity should be positive….

49. Hikarinara1217 says:

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

50. Ajax Holmes says:

the "-" stands for going down

51. tarfari Herry says:

Can't you use the timeless equation

52. ALI MUNTH says:

LOLOLOLOLOLOL

53. denny amiruddin says:

(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.

55. Miguel de Carvalho says:

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

56. coolawesomewowcrazy says:

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

57. JayTheDon says:

I get it now. thank you so much.

58. Marc Gatus says:

thanks it helps me to physics thumps up 🙂 👍

59. Laurelindo says:

@aiman fitri
a is a general notation for "acceleration", whereas g is specifically used for the acceleration of gravity.
You can use a instead of g, as long as it is clear that you mean the acceleration of gravity.
Alternatively, you can write a with the subscript g.

60. Haru Haru says:

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

61. HELL IsREALJesusSaves says:

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62. shubham kokane says:

superb !! bro keep it up ! you saved me…. now i am a scholar in physics….. ^_^

63. M Khan says:

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..

64. WokDaHoe says:

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

65. S Clair says:

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

66. Nina Duran says:

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

67. Cinder says:

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

68. Dyno says:

"When you learned about average velocity in elementary school"

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

69. SuperKoala CannonGawd/PvP Legend says:

whos watching this for ms.cole?? lmaaoo

70. Mynor says:

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

71. Niloy Das says:

wouldnt it be displacemnt instead of distance

72. Eitan Avni-Heller says:

Acceleration due to gravity is not 10. It is 9.8

73. Harshita Kanal says:

Shouldn't it be displacement?

74. Deep Gill says:

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

75. Mahabub Alma says:

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

76. Marianne says:

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

77. Kea Morshed says:

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

78. Krypto Knight says:

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 👌

79. RyZe_Android 17 says:

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! 🙏😊

80. iOwnLegacy says:

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

81. Nikhil Rauniyar says:

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

82. Mathilda Rose says:

Physics exam tomorrow, all-nighter let's gooo

83. INV Kristyler says:

I hope you put the standard formulas

84. Julius Soldan says:

85. Hanif Ameen says:

How does final velocity – initial velocity= acceleration . time

86. Lauren Packwood says:

You, Sir, are amazing. That is all.

87. jaison antony says:

sal seems sad in this video.

88. Pulkit Sinha says:

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

89. Jeeveash Kumar says:

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

90. EssentiallyFalse says:

I love learning in 240p quality.

91. suvansh gupta says:

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

Who else feels the same???

92. Anjali Shrivastava says:

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

93. Pide Tods says:

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?

95. Jeremy says:

What if ur given mass?

96. Deniz Mert Kunduş says:

Really amazing

97. Just Do It says:

We should experiment falling of from a cliff

98. Asif Khan says:

ugh the video quality…

99. Vijaya Valli says:

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

100. Gaming Tube says:

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