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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.

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