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

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

October 10, 2019 92 By Kody Olson


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.