Bo: Hey guys. Billy: Hey Bo. Bobby: Hi Bo. ♫ (lyrics) Flipping Physics ♫ Mr. P.: Ladies and

gentlepeople, the bell has rung therefore class has begun

therefore you should be seated in your seats, ready and excited to watch me drop a medicine ball. Billy: Seated! Bobby: Ready! Bo: Excited, woohoo. Mr. P.: We’ve already learned

the basics of free fall and today we’re going to

do an example problem. Because you are so excited, Bo,

you get to read the problem. Bo: Ok, thanks. Mr. P drops a medicine

ball from a height of two point zero meters above the ground. Part A: What is the velocity of the ball right before it strikes the ground? Part B: How long did the ball fall? Mr. P.: This is our picture,

the medicine ball starts out two meters off the ground

and it is in the vacuum that you can’t breathe, of course. Billy could you please translate? Billy: Uhhmm- Bo: Hold it, I have a question. Is that a meters-, no a

two-meter stick in the picture? Mr. P.: Yes that’s how I

knew that the medicine ball was two point zero meters off the ground. Bo: Then why aren’t you

holding the medicine ball at the top of the two-meter stick? Clearly, in the video you

can see the medicine ball is above the two-meter stick. Billy: Yeah! Mr. P.: Ah, this is called parallax. It’s an important question. It happens because the video

camera is actually below the top of the two-meter stick and the medicine ball is in

front of the meter stick. Therefore the medicine

ball appears to be above the top of the two-meter stick. Again, it’s called parallax. You can see, we actually

have the same issue when the medicine ball is at the

very bottom, because you know that the medicine

ball and the two-meter stick are both on the ground,

however it appears as though the medicine ball is actually below the bottom of the two-meter stick. Again, called parallax. But you can believe, I

did hold the medicine ball at the very top of the two-meter stick. Bo, thank you very much

for bringing that up. Billy, could you please translate now? Billy: Let’s see, we

know the displacement. Delta X equals two point zero meters. Mr. P.: True, we have been

using an X for displacement, but because this displacement

is actually in the Y direction we’re going

to use delta Y instead. Mr. P.: Billy, please keep going. Billy: The final velocity equals zero. Mr. P.: Ah yes, I agree,

the final velocity of the medicine ball at the end of the

video clip is equal to zero. That’s definitely true. However, the final velocity

in our problem is not zero. Who could tell me why? Bo: Because it’s touching the ground and no longer in free fall. Billy: Right, in order to

be in free fall it can’t be touching any other objects. Oh, and we’re looking

for the final velocity right before it hits the ground. So part A, velocity final equals

question mark and we might as well add for part B change

in time equals question mark. Mr. P.: Yes. As soon as the medicine ball

strikes the ground it is no longer an object in free

fall and we don’t know the physics yet to work with that. So what we’re trying to

find is the final velocity right before it strikes

the ground and that is the final velocity in the Y direction. So that equals question

mark, that’s part A. And for part B we are trying

to find the change in time while it’s flying through the

vacuum that you can’t breathe. Billy please keep going. Billy: It’s in free fall

so the acceleration in the Y direction is negative

nine point eight one meters per second squared. Mr. P.: Therefore because

the acceleration is constant the acceleration equals

a number, we can use the uniformly accelerated

motion equations. However we need to know three

of our UAM variables before we can solve for the other two. And I only see two known

UAM variables right here. So who can tell me what

is the third UAM variable, that we already know? Bobby: We know the initial

velocity is zero because you dropped it, you can

see in in the video that it isn’t moving right before you drop it. Mr. P.: True, that word drop

is a very important one it tells us that the medicine

ball isn’t moving at the very beginning therefore

the initial velocity, and I’ve added a Y to identify

that it’s the initial velocity in the Y direction, that is equal to zero. Now we know three of our UAM variables, we can solve for the other two. Yes Billy, you have a question? Billy: Don’t we need to know

the mass of the medicine ball? Bobby: Actually no, remember

that mass doesn’t matter. Any object in free fall will

have the same acceleration. Billy: Yeah, that’s right, thank you. Bobby: You’re welcome. Mr. P.: Thanks Bobby and

Bobby could you please tell me how we are going to solve for the final velocity in the Y direction? Bobby: We just need to find

the right UAM equation. Let’s see, we could use

final velocity equals initial velocity plus acceleration

times the change in time. That actually won’t work. Mr. P.: That is correct Bobby, why can’t we use that equation? Bobby: Because we don’t know

the final velocity or the change in time and we can’t solve one equation with two unknowns. Let’s use final velocity squared equals initial velocity squared plus two times the acceleration times the displacement. Mr. P.: Very close, remember

everything is in the Y direction so we’re going to add subscripts

of Y and we’re going to change the displacement to the displacement in the Y direction. I do understand that

everything in this problem is in the Y direction and many

of you are going to want to eschew the Y because it takes so long to write down all those Ys. But please do not eschew the Y. I’m trying to get you

ready for projectile motion and in projectile motion

we have things moving in both the X and Y direction

at the same time and therefore we have to use

subscripts of X and Y. And you have to keep them separated. And it’s just a good habit

to get into right now so that you’re more used to it when we do get to projectile motion, please. Bo: I won’t eschew the Y. Mr. P.: Good, I’ll plug

in the numbers now. The initial velocity is equal

to zero which we square which just makes it still

zero plus two times the acceleration Y direction which is negative nine point eight one

times the displacement in the Y direction which is two. Taking the square root of

both sides we get to the final velocity of the Y direction

is equal to the square root of two times negative nine

point eight one times two. And what do we get? Billy: Uhm, it’s six point

two six four one eight. Mr. P.: Sorry, not possible. Billy: Yeah well, you

can’t take the square root of a negative number so

I just made it positive. Mr.P.: You can’t do that, you can’t just make a number positive. Anytime that you’re trying take

a square root of a negative number in this class it means that you have done something wrong. So we did something wrong here. But remember, I don’t

usually correct mistakes immediately because it is

important that you make mistakes and see what happens because

of those mistakes so that you can recognize them in the future. So, we made a mistake, where is it? Bo: Displacement has both

magnitude and direction. But displacement should be

negative two point zero meters. Mr. P.: Yep, the displacement

in the Y direction is negative two point zero meters. There are two ways of looking

at this, one of them is that the displacement in the

Y direction is going to be the Y position final minus

the Y position initial. Therefore zero minus two

because it starts out two meters above the ground,

thats it’s initial position, and it’s final position

is at the ground, or zero. Therefore zero minus two is

negative two point zero meters. You could also simply say

because we’ve defined down as negative, and we know that the

medicine ball is moving down, therefore the displacement in the Y direction is negative

two point zero meters. Where that negative

means it is moving down. So we carried the negative

through our equations and we’re now taking the square

root of a positive number Billy, what do we get? Billy: Our final velocity is

six point two six four one eight, which rounds to six

point three meters per second with two sig figs. Mr. P.: Now that we

have our final velocity in the Y direction, six point

three meters per second, Bobby could you please solve for part B, the change in time? Bobby: Actually, now that we

know four of our UAM equations there are three equations

we could use, right? Mr. P.: Very perspicacious of you, Bobby and you are correct. Bobby: Great, I’ll pick

velocity final equals- Mr. P.: Bobby please add

all the Ys for direction! Bobby: Velocity final in

the Y direction equals velocity initial Y plus the acceleration in the Y direction

times the change of time so we get six point two six four one eight equals zero plus negative

nine point eight one times the change in time divide both sides by

negative nine point eight one and you get the change in

time equals six point two six four one eight divided by

negative nine point eight one which is… Bo: negative zero point six

three eight five five one which rounds to, with two sig figs, negative zero point six four seconds. Hold up, that doesn’t make any sense, we can’t have a negative time. Mr. P.: Yep, negative time

means we made another mistake. Bobby: Woah?! Billy: Oh man! [Bobby sighs] Mr. P.: You have to be

smarter than your calculator. Billy: Oh, our final velocity

should be negative because the medicine ball is going down! Bobby and Bo: Aahhhh! Mr. P.: Yes, anytime you take

the square root of a number you need to be smarter

than your calculator and realize your calculator lies to you. You calculator only gives

you the positive answer. However, YOU, need to think

to yourself: Aha, it’s a square root, the answer could

be positive or negative. And we know that final velocity

is the Y direction must be negative because the

medicine ball is going down. Please, be smarter than your calculator. As you can see the negative

carries all the way through and we end up having a positive

change in time which is correct. One other little thing,

some of you are going to get overzealous with your final

velocity and really want to tell me it’s in the down

direction and therefore you’re going to write it like this. Negative six point three

meters per second down. Class, what direction is negative six point three meters per second down? Bobby and Bo: Up. Mr. P.: Yes, negative six point

three meters per second down is actually up, which is sadly wrong. You could write for your final

velocity in the Y direction negative six point three meters

per second or you can write six point three meters per second down, both are equally correct. However, when you’re working

with a problem you need to use the negative for your variable, got it? Billy: Clear as a bell. Bobby: What?! Billy: Clear as a bell! Mr. P.: You know what as long

as we have the video here let’s take a moment and make

sure that the problem we did matches the reality of the video. Now, I filmed the video at

a 120 frames per second so I can show it to you now

at 30 frames per second and have it at 25 percent

normal speed, let’s take a look. Will you look at that, the physics works. (Mr. P. starts singing) The physics works, the

physics works uh-huh-uh-huh, the physics works, the

physics works uh-huh-uh-huh, the physics works- Bo: That’s just awkward. Bobby: Yeah. Mr. P.: The physics works uh-huh-uh-huh. (awkward laughter) Alright, let’s do a quick review. We listed everything that we

knew, we identified that we had uniformly accelerated

motion because the acceleration was equal to a number, it’s

free fall, that’s always true. We identified three UAM

variables, we solved for the other two using UAM equations. The key here is to understand the three common mistakes that we made. First off, the final velocity

in the Y direction is not equal to zero because that is only true after it strikes the

ground, it is no longer in free fall when that happens. The second was that the

displacement in the Y direction needs to be negative because

the object is going down. And the third is anytime

you take the square root of a number, you need to be

smarter than your calculator and decide whether the answer

is positive or negative. That’s what we we did. Thank you very much for

learning with me today class. I enjoyed learning with you. (Mr. P. starts singing) The physics works, the

physics works uh-huh-uh-huh, the physics works, the

physics works uh-huh-uh-huh. (awkward laughter) That’s it.