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# Projectile motion (part 5)

Video transcript

Welcome back. Let's continue doing projectile
motion problems. I think this video will be
especially entertaining, because I will teach you a game
that you can play with a friend, and it's called let's
see how fast and how high I can throw this ball. You'd be surprised--
it's actually quite a stimulating game. Let me just write
down everything we've learned so far. Change in distance is
equal to the average velocity times time. We know that change in
velocity is equal to acceleration times time. We also know that average
velocity is equal to the final velocity plus the initial
velocity over 2. We know the change in velocity,
of course, is equal to the final velocity minus
the initial velocity. This should hopefully be
intuitive to you, because it's just how fast you're going at
the end, minus how fast you're going at the beginning,
divided-- oh no, no division, it's just that I got
stuck in a pattern. It's just vf minus
vi, of course. You probably already knew this
before you even stumbled upon my videos, but-- the two
non-intuitive ones that we've learned, they're really just
derived from what I've just written up here. If you ever forget them, you
should try to derive them. Actually, you should try to
derive them, even if you don't forget them, so that you
when you do forget it, you can derive it. It's change in distance-- let
me change it to lowercase d, just to confuse you-- is equal
to the initial velocity times time plus at squared over 2,
and that's one of what I'll call the non-intuitive
formulas. The other one is the final
velocity squared is equal to the initial velocity
squared plus 2ad. We've derived all of these,
and I encourage you to try rederive them. But using these two formulas
you can play my fun game of how fast and how high did
I throw this ball? All you need is your arm, a
ball, a stopwatch, and maybe some friends to watch
you throw the ball. So how do we play this game? We take a ball, and we throw
it as high as we can. We see how long does the
ball stay in the air? What do we know? We know the time for the ball
to leave your hand, to essentially leave the ground and
come back to the ground. We are given time, and
what else do we know? We know acceleration-- we know
acceleration is this minus 10 meters per second. If you're actually playing
this game for money, or something, you would probably
want to use a more accurate acceleration-- you could look
it up on Wikipedia. It's minus 9.81 meters
per second squared. Do we know the change
in distance? At first, you're saying-- Sal,
I don't know how high this ball went, but we're talking
about the change in distance over the entire time. It starts at the ground--
essentially at the ground, because I'm assuming that you're
not 100 feet tall, and so you're essentially at the
ground-- so it starts at the ground, and it ends of the
ground, so the change in distance of delta d is 0. It starts with at the ground
and ends at the ground. This is interesting-- this is a
vector quantity, because the direction matters. If I told you how far did the
ball travel, then you'd have to look at its path, and say
how high did it go, and how high would it come back? Actually, if you want to be
really exact, the change in distance would be the height
from your hand when the ball left your hand, to the ground--
so, if you're 6 feet tall, or 2 meters tall, the
change in distance would actually be minus 2 meters,
but we're not going to do that, because that would just be
too much, but you could do it if there's ever a close call
between you and a friend, and you're betting for money. You're given these things,
and we want to figure out a couple of things. The first thing I want to figure
out is how fast did I throw the ball, because that's
what's interesting-- that would be a pure test
of testosterone. How fast? I want to figure out vi--
vi equals question mark. Which of these formulas
can be used? Actually, I'm going do it first
with the formulas, and then I'm going to show you
almost an easier way to do it, where it's more intuitive. I want to show you that these
formulas can be used for fun with your friends. We know time, we know
acceleration, we know the change in distance, so we could
just solve for vi-- let's do that. In this situation, change in
distance is 0-- let me change colors again just to change
colors-- so change in distance is 0 is equal to
vi times time. Let me put the g in for here,
so it's minus 10 meters per second squared divided by 2,
and it's minus 5 meters per second squared-- so it's
minus 5t squared. All I did it is that I took
minus 10 meters per second squared for a, divided it
by 2, and that's how I got the minus 5. If you used 9.81 or whatever,
this would be 4.905 something something. Anyway, let's get back
to the problem. If we wanted to solve this
equation for vi, what could we do? This is pretty interesting,
because we could factor a t out. What's cool about these physics
equations is that everything we do actually has
kind of a real reasoning behind it in the real world,
so let me actually flip the sides, and factor out a t, just
to make it confusing. I get t times vi minus
5t is equal to 0. All I did is that I factored out
a t, and I could do this-- I didn't have to use a quadratic
equation, or factor, because there wasn't a
constant term here. So I have this expression, and
if I were to solve it, assuming that you know vi is
some positive number, I know that there's two times where
this equation is true. Either t equals 0, or this term
equals 0-- vi minus 5t is equal to 0, or since I'm solving
for velocity, we know that vi is equal to 5t. That's interesting. So what does this say? If we knew the velocity, we
could solve it the other way. We could say that t is equal to
vi divided by 5-- these are the same things, just solving
for a different variable. But that's cool, because there
are two times when the change in distance is zero-- at time
equals 0, of course, the change in distance is zero,
because I haven't thrown the ball yet, and then, at a later
time, or my initial velocity divided by 5, it'll also
hit the ground again. Those are the two times
that the change in distance is zero. That's pretty cool. This isn't just math--
everything we're doing in math has kind of an application
in the real world. We've solved our equation--
vi is equal to 5t. So, if you and a friend go
outside and throw a ball, and you try to throw it straight
up--0 although we'll learn when we do the two dimensional
projectile motion that it actually doesn't matter if you
have a little bit of an angle on it, because the vertical
motion and the horizontal motions are actually
independent, or can be viewed as independent from each other--
this velocity you're going to get if you play this
game is going to be just the component of your velocity
that goes straight up. I know that might be a little
confusing, and hopefully it will make a little more sense in
a couple of videos from now when I teach you vectors. If you were to throw this ball
straight up, and time when it hits the ground, then this
velocity would literally the speed-- actually, the
velocity-- at which you throw the ball. So what would it be? If I threw a ball, and it took
two seconds to go up hit the ground, then I could
use this formula. This is actually 5
meters per second squared times t seconds. If it took 2 seconds-- so if
t is equal to 2-- then my initial velocity is equal
to 10 meters per second. You could convert that to
miles per hour-- we've actually done that in
previous videos. If you throw a ball that stays
up in the air for 10 seconds, then you threw it at 50 meters
per second, which is extremely, extremely fast.
Hopefully, I've taught you a little bit about a fun game. In the next video, I'll show
you how to figure out-- how high did the ball go? I'll see you soon.