Projectile motion (part 5)

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Projectile motion (part 5) : How fast was the ball that you threw upwards?

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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.