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.

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Dennis Veasley

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