Today, we're going
to take it quite easy. I also have to take it
a little easy because my voice may be petering
out, if I'm not careful. We're going to apply today
what we have learned, so there is nothing new
but its applications. And that's important-- things
that... you can let it sink in. We have here a trajectory of a golf ball
or a tennis ball in 26.100. We shoot it up
at an angle alpha. The horizontal component
in the x direction is v zero cosine alpha and the vertical component
is v zero sine alpha. It reaches
the highest point at P and it returns to the ground
at point S. This is
the increasing y direction and this is
the increasing x direction. We're going to use,
very heavily, the equations that you see here
that are so familiar with us. These are the one-dimensional
equations in x direction when there is no acceleration and the one-dimensional
equations in the y direction where there is acceleration. In order to use these equations we need all these constants-- x zero, v zero x and v zero y. We have seen those last time. I choose for x zero...
I choose zero arbitrarily. Also for y zero. The velocity in the x direction
will never change. This v zero x will always remain
v zero cosine alpha. The velocity in the y direction,
however, in the beginning at t equals zero
is v zero sine alpha. And that one will change,
because there is here this t and that's why the velocity
is going to change. This t will do it. And the acceleration
in the y direction-- since this is
increasing value of y-- is going to be negative 9.8. Since I call always 9.8 plus... since I always call g
"plus 9.8," this is minus g. I now want to ask first the question that you may
never have seen answered: what is the shape of this? Well, we can go to
equation number three there and we can write down
this equation number three: That y, as a function of time,
equals v zero yt so it is v zero sine alpha
times t minus one-half gt squared. That's the equation in y. I go to equation number one and I write down x--
at any moment in time-- equals v zero z times t so that is
v zero cosine alpha times t. Now I eliminate t, and the best way to do that
is to do it here-- to write for t, x divided
by v zero cosine alpha. Now I can drop
all subindexes t because we're now going
to see x versus y. We're going to eliminate t. So this time here, I'm going to
substitute in here and in there and so I'm going
to get y equals... There's a v zero here and there's a v zero there
that cancels. There's a sine alpha here
and a cosine alpha there that makes it a tangent
of alpha. And then I have here the x and I get minus one-half g
times this squared-- x squared divided by v zero
cosine alpha squared. And now look very carefully. Y is a constant times x minus another constant
times x squared. That is a parabola. It's a second-order equation
in x, and is a parabola and a parabola has this shape. So you so see now,
by eliminating the time that we have here a parabola. Now I want to massage
this quite a bit further today. I would like to know at what time the object here comes to a halt
to its highest point. It comes to a halt
in the y direction. It comes to a highest point and I want to know
how high that is. Well, the best way to do
is to go to equation four and you say, to equation four, "When are you zero?" Because that is the moment that the velocity in the y direction becomes zero. It must be
at its highest point, then. So in order to find, for us, the position
of the highest point P we first ask ourselves the question
from equation number four: when is the velocity
in the y direction zero? And that then becomes
v zero y which is v zero sine alpha
minus gt and out pops that t at point P is going to be v zero sine alpha
divided by g. That's the time that it takes for the object
to reach the highest point. Where is it, then? What is the highest point
above the ground? Well, now we have to go
to equation number three and you have to substitute
this time in there so that highest point h, which is y at the time t of P
equals v zero yt-- that is v zero
times the sine of alpha. But you have to multiply
it by this time and so I get
another v zero sine alpha and I get a g here
minus gt mi... oh, no, no, this equa...
minus half dt squared. Minus one half g
times this one squared. which is v zero
sine alpha squared, divided by g...
divided by g squared because there is a g here,
you see? So you square the whole thing
if it's t squared. You lose one g and you will find, then,
that the highest point-- let's write it down here so that
we don't block that blackboard-- the highest point in the sky equals v zero sine alpha squared
divided by 2g. That is the highest point. Let's give that some color because we may want
to keep that. Is it reasonable that the point,
the highest point in the sky gets higher when v zero
is higher? Of course. If I shoot it up
at a higher speed of course it will get higher. So that's completely intuitive
that v zero is upstairs. If I increase the angle
from a small angle to larger and larger and larger is it reasonable
that it will get higher? Of course. You all feel in your stomach that the highest possible value
you can get is when you make alpha
90 degrees for a given velocity. That's the highest it will go
in the sky. So clearly, this is
also very pleasing. If you did the experiment
on the moon with the same initial speed,
it will go much higher so you are also happy to see
that this g here is downstairs. So that makes sense. At what time will the object
be at point S? Now, there are two ways
that you can do that. You either go to this equation,
number three and you ask equation number
three, "When are you zero?" It will give you two answers. It will say, "I am zero here
at this time" and "I am zero at that time." And those are the two times
that you want and this is the one you pick. That's perfectly fine. I think there's a faster way to
do it, and that's the following. This is a parabola, so it's completely symmetric
about the vertical, about P. So to climb up from O to P must
take the same amount of time as to go down from P to S and so I claim
that the time to reach point S must be twice the time
to reach point P and therefore it's going to be two v zero sine alpha
divided by g. But now we want to look again whether the v zeros and the sine
alphas have the right place. Indeed, if I increase the speed, I would expect it to take longer
before it reaches S. If I give it a larger speed,
it will come out farther and obviously,
the time will take longer. If I do it at a higher angle,
it will also take longer and if I do it on the moon,
it will also take longer. So this makes sense-- these equations are pleasing
in terms of their rate that v zero and sine alpha
appear in the equations. But now comes
an important point which I am going to use
throughout this lecture. I want to know what OS is. The distance OS... I shoot it up
and it hits the floor again What is that distance
that it travels? Well, for that,
I need equation number one. It is v zero x times the time and v zero x
is v zero cosine alpha. We got v zero cosine alpha
times the time to hit it-- that is two v zero sine alpha. So I get a two here, I get a
sine alpha, and I get a g here and I have another v zero there,
and so the answer is v zero squared times
the sine of the double angle-- remember, two cosine
alpha sine alpha is the sine of two alpha--
divided by g. And this is OS, and I'm going
to need this a lot. This reminds me
not to remove it. Now, I sort of wonder,
and you should too why is it that the highest point
in the sky has a v zero squared and why is
the farthest point also... why does it also have
a v zero squared? There must be a way
that you can reason that. Why is it not just v zero? Why is it v zero squared? Well, I'll let you argue
about the highest points, and I'll give you a good reason
for the distance, OS. Don't look at the equations. You simply... Think for a change. Don't look at the equations. I double the speed. If I double the speed,
then it's quite reasonable that the time that it takes for the object to reach
the ground will double, but while the time
that it flies has doubled the horizontal velocity
has also doubled. And so the distance that it will
travel in horizontal direction is four times that-- twice
because the time has doubled and another factor of two because the horizontal component
has also doubled. So that's why you see
v zero squared there-- completely pleasing. This tells you immediately
that the... if you want to throw a ball
as far as possible-- people who play baseball
know that-- you should do it at 45 degrees. Because if you throw it
at 45 degrees then this angle, 90 degrees,
and that is one. Of course, in reality, a baseball player
knows better. They give effect to the ball,
they deal with air drag they spin the ball, and then
these equations are not valid. This is only in case we deal
with... with vacuum. I now would like to test some of the results
that we have here... we have worked out here. I am going to shoot a pellet...
a metal ball. I'm going to shoot it
at various angles: 30 degrees, 60 degrees,
45 degrees and I'm going to make
a prediction if I shoot it up from there,
where it will hit the table. A measurement is meaningless without knowing
the uncertainties. So that's the first thing
we have to deal with. The first thing I want to know is what is the velocity
of this bullet when it comes out of the spring and does it vary if I do it
three, five, six times in a row? It's not a $20,000 spring gun,
so it is likely to vary. And the way I am going to do
that is as follows. If I shoot an object
vertically up-- that is, the maximum value
that it can go-- and with an alpha
equals 90 degrees then the sine of alpha is one and the height is v zero squared
divided by two g. In other words,
if I measure the height if I shoot it up vertically and you can measure
that for me-- you will see how I am asking
you to do that-- then we can calculate
v zero squared. So the first thing I want to do
is to shoot it up vertically and how are you going
to help me to calculate... to tell me how high it is? That comes easier
than you think. The top part...
oh, we'll remove this. The top part of this stick
is three meters, the top mark. That very top mark
is three meters and all I want you to tell me whether it is yay much above
or yay much below and then we'll estimate
that yay much and then we'll make a guess. And I'll do it twice. So if you are ready? Make sure that you
can distinguish between above and below--
it makes a big difference, yeah? Okay? Three, two, one, zero. (ball whooshes) Okay, was it higher or lower? CLASS:
Higher. LEWIN:
How much? This much? Do we agree? Let's say
five centimeters, right? We're going to allow
for an uncertainty. I'll do it again. I want to see
how well it reproduces. Three, two, one, zero. (ball whooshes) (students shout out answers) Lower? STUDENT:
Higher. LEWIN:
Higher! So it was 10 centimeters,
five centimeters higher so we'll take seven. So we'll make seven and we'll have to allow
for an uncertainty. So h max... is about 3.07. I've done this
this morning 20 times and there were times
that the heights differed by more than 10 centimeters,
sometimes even 15 centimeters. I therefore would feel
most comfortable if you allow me an uncertainty of 15 centimeters
in that height. Remember, once we start shooting
at 30 degrees, there is no way that we can evaluate
the velocity anymore. We have to just take
this value at face value. This is the way we've measured
v zero, and that's it. This is a five percent error,
five percent. So what now is v zero squared? Well, that's easy
to calculate now. V zero squared equals
3.07 times 2 times 9.8... Oh, my calculator was off;
that's a detail. Um, 3.07 times 2 times 9.8--
that is 60.17. I'd like you to check that. 60.17 plus an error
of five percent. That is an error of three so you might as well make
this 60.2. Would you please confirm that,
that I didn't make a mistake? 3.07 is h max-- I multiplied
by two, by 9.9, and 60.2 There's a five percent error and a five percent error
is indeed three. This is meters squared
per second squared. I don't care what v zero is because if we are going
to measure OS all I need is v zero squared. And if you at home
are going to calculate what the height will be
at the various angles all you need is v zero squared. So I am not even interested
in v zero. I'll just stick
to v zero squared and v zero squared will have
exactly the same uncertainty. It will have an uncertainty
of five percent because it comes immediately
from h. We are not going to change that. Okay, so, so much for the
uncertainty in v zero squared. Now I'm going to set
the angle at 45 degrees but how accurately
can I do that? I don't think I can do that
any better than one degree. I'll try to do the best I can. But I can't really guarantee you
that I'm accurate to one degree. So now comes the question what happens with
the sine of two alpha because we're going
to measure OS? What happens with
the sine of two alpha? The sine of 90 degrees
is 1.0000. But what would be
the sine of 88 degrees? That is the value
that I cannot exclude if I'm off by one degree. And that value is 0.9994. That is so close to one that it is only off
by 0.6%... 0.06 percent. And that is so low, compared
to five percent, forget it. Forget the error in alpha. We can completely forget it. There's a reason for that. When alpha is 45 degrees,
then 2 alpha is 90 degrees and the sine curve goes
like this at 90 degrees. It's almost flat here. So even if you're off
an angle by a little bit you're still very close to one--
that's the reason. So all we have to worry about is the uncertainty
in v zero squared. And so now comes
my big prediction. I'm going to make
a prediction now: for 45 degrees,
OS equals v zero squared. We have that.
That is 60.2. And we have the sine
of two alpha is one and we divide by 9.8. That is 6.14 meters with an uncertainty
of five percent, right? Because that is the uncertainty
in v zero squared and so there is an uncertainty of 30 centimeters,
actually 31 centimeters. This is my prediction
for an angle of 45 degrees. This will only hold
if there is no air drag or if the air drag
is negligible, and of course,
equally important, that that spring gun--
when the ball comes out-- that the velocity squared
is indeed within the range that we have assumed and that it doesn't have
bad days and good days. There's no way
I can check that anymore. All right, so we're going
to mark the .614. This is one meter, two meters,
three meters, four meters five meters, six meters, 6.14. One four... 14 centimeters. Boy, God, it's
all the way here. And then I allow for an error
of about 30 centimeters. Did I do that right? That is correct, 45 degrees
with 30 centimeter uncertainty. That is all the way up to here. And then the next one,
roughly 30 centimeters. So that's where, if our
calculations make sense that's where
the ball should hit. Now I would like you
to come here, if you don't mind. And stand here, and the moment
that that ball hits... (whooshes) point your finger at it. Don't do it before I shoot,
but just after I shoot. And then we'll hope
for the best, yeah? Okay. You're not nervous, right? Where... what happened
with that ball that I had? Did I put it in my pocket? Oh, it's here. Thank you. So I'm going to set it now, to
the best I can, at 45 degrees. And so I can never shoot it any
further, this is the angle... this is
the maximum possible distance. You're ready? You are? Don't look at me, look there. It goes fast. Three, two, one, zero. (gun clicks) LEWIN:
Put your finger there! (class laughs) Isn't that fantastic? Isn't that amazing? Do you see now
how important it is that you have uncertainties
in your measurements? In high school, you would have
said it has to hit there. Boom, man, it has an error. (class laughs ) And the error has to
be taken into account. Where's my ball, by the way? Boy... Oh, I have it, ooh, here. (grunts) Okay, you can sit down now. You did great. It worked just because
you were there. (class applauds) So now I wonder what happens if
I fire the ball at 30 degrees? Well if I do it at 30 degrees, then
my v zero squared is the same. I don't have to worry
about that, I hope. However, I cannot be certain
about the angle to better than one degree. So you will say, "Well,
come on now, don't be decadent. "I mean, here we had an error
of only 0.06 percent "because of this one degree...
possible one degree offset. Let's just ignore
this error, too." Ooh... that is risky. That is risky, now
because the sine of 60 degrees-- that's what you deal with-- the sine of 60 degrees,
I think, is 0.866. That's right. But the sine of 58 degrees which is possible
if I'm one degree under is 0.848, and that is
substantially lower. And therefore I must allow for an uncertainty
in the sine of the angle by roughly, oh, maybe
something like 17 or 18 units-- 0.0... what is this difference? 0.018. If you want to check
what the sine of 62 degrees is you will see that that is about
this much higher than this so we must allow for this error. And that is an error which is
by no means negligible anymore. There's no point here. That is an error which is
18 divided by 866. That is a two percent error. So now we're dealing,
all of a sudden with a two percent error
in the sine of two alpha even though there is only
one degree of uncertainty in the angle. And the reason is
that a sine curve is like so. So a small angle change here
makes no difference but a small angle change here
makes a lot of difference. And that's the reason and you can see that it is
the slope of the sine curve that gives you
a much larger error if you are off
by a teeny-weeny, little bit. So now we are ready
to make a prediction. So here comes the prediction. OS now for 30 degrees... So I have to go through v zero
squared-- I have that, 60.2 and then I have to multiply
by the sine of two alpha that is, the sine of 60...
multiply... and then I think
I have to divide by g. That is right, 9.8,
and I find 5.31, plus or minus. Now, two percent error
in the sine of two alpha five percent error
in v zero squared that gives me-- I can't
help it-- a seven percent error. So I have a seven percent
uncertainty, multiply by .07 so I cannot trust this
any better than 37 centimeters. And so now I'm going to put
the markers out at five meters
and 31 centimeters. This is the five-meter mark,
31 centimeters and I allow for 37 centimeters
on either side. It's here... and 37 centimeters
on this side, that is here. I'm going to set the angle
to 30 degrees. Yeah, yeah, come here. We need woman power here. (class laughs) Stay here. Stay out of the fire line,
and when that ball hits you jump on it,
yeah, you jump on it. Okay, everything under control Five... make sure
I marked it right, five meters, three one,
that looks about okay. And now I have to change
the angle to 30 degrees. (grunts) Okay, this is as close
as I can do it. Okay, you're ready? You are? Three, two, one... (ball whooshes) (Lewin shouts: "Ah!") Lewin:
Hit the jackpot! (class cheers and applauds) Incredible. What you can argue now,
and successfully-- you could say
perhaps you have been a little too conservative on
your errors, and I admit that. But believe me when I did this morning
this five, six times that the error in v zero
was quite substantial. The high differences were
sometimes 15 centimeters and so I had no choice. But it looked like we were
a bit on the conservative side. Suppose, now, I repeated
this experiment at 60 degrees. What will change? What will change? Will OS change? No, because the sine
of 120 degrees is the same as the sine
of 60 degrees. You have two alpha here. The sine of 60 is the same
as the sine of 120. If you allow for an uncertainty
of one degree on either side you will find
exactly these same numbers because of the symmetry
of the sine curve. So again, you are off
by two percent-- no difference. And so this prediction
is unchanged. However, I want
to ask you one question to see whether you are
half-alert or half-asleep. At 30 degrees, you saw this. At 60 degrees, giving it
the same initial speed you will see this. It will have to hit here within the uncertainty
of our measurements at the same location. It will, of course,
go much higher. You can calculate that because
you will have to use this for the equation for the height and that goes
with the sine of alpha. The sine of alpha for 60 degrees
is way higher than 30 degrees. But now comes a question-- will this trajectory
take longer than this one or will they take
the same amount of time? Who is for
the same amount of time? Who is for longer? Who is for shorter? Okay. I am happy
with one set of fingers and unhappy
with another set of fingers. What is the horizontal velocity of the golf ball
when it takes off? If the velocity is the same
in both cases can't you see that
this horizontal component is way larger
than this horizontal component? And if they travel
the same distance then this tripmust take longer because it's the horizontal
component in the x direction that determines how long it will
take to go from here to there. Suppose I shot it straight up. How long do you think
it will take to hit here? It has no horizontal component,
all right? So think about this--
this trajectory will take longer but it ends up
at the same point. All right, number three. Can you come here? So I don't have
to change these markers. They're all perfect provided that nature is
willing to reproduce itself. 60 degrees... So he'll go way higher, but if all goes well, it should
hit within the same marks. Ready? Three, two, one, zero. (ball whooshes) (ball whacks table) LEWIN:
Here, man. Yeah, here, right? Thank you very much. Wonderful! Look! Maybe my uncertainties
were not so dumb. We were just lucky here
with the jackpot. It's comfortably within
the error, but close to this one so I am quite happy that I took
the uncertainties the way I did. Can we recover the ball? Did someone see it take off? Oh, yeah. All right, now we will enter a
different part of this lecture. which is actually
a very... a very sad part. You know that people in Africa
shoot monkeys. There is a monkey here
in a tree-- very happy. (class laughs) LEWIN:
And here is a hunter... who never took 801 and he has a gun,
which is a golf ball gun and he aims that gun
right at the monkey. He shoots it with a certain
velocity, the golf ball. Let this be speed v zero,
so the horizontal component-- you're going to see that
again and again-- v zero cosine alpha and the vertical component
equals v zero sine alpha. Let this be
my increasing value of y and this be
my increasing value of x. This golf ball...
this... this golf gun, is really not first-class,
thank goodness. And so when the hunter shoots
this golf ball, this happens. And it ends up here at point p. Lucky monkey, so far. Now, it is very tragic but true that when the monkey sees the
flash of the gun, it lets go. (class laughs) And now comes the question-- is the monkey safe or is this
the last day of the monkey? (class laughs) I ask the following question. This would be trajectory
with no gravity and this is the trajectory
with gravity. We can both agree on that. At a certain moment, t1, let us assume
that the golf ball would have been here
without gravity. Then I know exactly
where it is with gravity. It must be exactly here because the x position, x t1,
is the same because the horizontal velocity
is the same. That's independent of
whether there is gravity or not. There is no acceleration
in the x direction. And so they are both exactly
at the same x position. What is this difference? Well, that is the difference between the equation with
gravity and without gravity. And y as a function of time-- you can look at equation number
three there if you can still see it-- equals v zero y, which is
v zero sine alpha times t minus one-half g t squared. Well, if there is no gravity,
this term doesn't exist so that's
this straight line. (whooshes) With gravity,
it's the same thing but you have to subtract this. Therefore, this distance
is one-half g t1 squared. That is this distance so this curve is lower
by this amount. Now comes the time that
the golf ball hits point p. When its position is x t2
and the time here is t2 that means if there
had been no gravity the golf ball
would have been there. They must have
the same position in x at this catastrophic moment. So what now is the distance between the monkey
and the golf ball-- the distance
between the two trajectories-- one trajectory, no gravity;
the other with gravity? This distance equals
one-half g t2 squared for that same reason. And we all know that if the monkey
at time t equals zero let go that in t2 seconds
it will have fallen exactly over a distance
one-half g t2 squared-- exactly. This couldn't be more tragic. And he will be killed. You may say, "Well, yeah,
but you have manipulated the speed of that gun
just so nicely." Oh, no. Oh, no. I can shoot that with a higher
speed at the same angle alpha and the trajectory would be this and the monkey
would be killed there. I can do it with a lower speed and the monkey
would be killed here. It's independent
of the speed of the bullet because always this part here--
it's always exactly the distance that the monkey falls
in that time. However,
if the speed is very low-- that it hits the ground before
the monkey hits the ground-- well, okay,
then the monkey is safe. So the only thing that is very,
very critical is alpha. It must be precisely aimed
at the monkey. If that's not the case,
then the monkey will be safe. Now before we will witness this
classic and rather tragic drama, I want to look at this from a somewhat
different point of view namely from the point
of view of the monkey. The monkey sits there,
looks at the gun and the golf ball comes
to the monkey. And I will put them both in
a room which is an elevator and the elevator is in free fall and they don't even know that. They both fall
with the acceleration g. Here is the monkey, free and here is the gun. The velocity of that bullet
is v zero. And so the monkey will see that
bullet come straight at him. There's no such thing as an arc. They both fall
in this falling grav... in this falling elevator. And so the bullet comes... (kisses) The monkey happens to be
a very intelligent monkey and the monkey says to himself,
"How long do I have to live?" And the monkey makes
the following calculation. (class laughs) LEWIN:
If this distance is d, and this
is h, then the monkey says "Aha, this is the square root
of d squared plus h squared." So from the monkey point
of view, the time for the kill will be the square root of d squared plus h squared
divided by v zero. That's how many seconds
he has to live. Well, you people
are also quite smart and you look at this diagram
and you said, "No, no way." If this distance is D then the speed to reach this
point is v zero cosine alpha. In other words, the time
that it takes for this object to reach this value of x... So, for 26.100 MIT students t kill equals D divided
by v zero cosine alpha. But what is the cosine of alpha? That is D divided
by the square root of D squared plus h squared. So I can replace
this cosine alpha by D divided by the square root. So I can replace
this cosine alpha by D divided by the square root. And you and the monkey
agree exactly on the amount of time
that the monkey has to live. It better be that way,
because this could not depend on which reference frame
you work in-- the falling reference frame or, for that matter
the reference frame of 26.100. The monkey will be placed at about three meters
above the table. We all know that it takes
about 0.8 seconds. We have done many experiments
at three meters. It takes about 0.8 seconds. So the whole thing
will go very fast. We are going to put
a monkey up there. I want you to first see
the trajectory of that golf ball before we bring the monkey in. It is already so painful
for this monkey. You don't want him
to pre-experience what's going to happen. So we will do this
in the absence of the monkey and we will let you...
I will let you see what roughly the trajectory
of that bullet will be. Three, two, one, zero. (gun claps) (ball bouncing) So it will hit somewhere here. That's that point p. So when you're going to see
the drama in action this is where
the monkey will reach when the two hit each other. Now you can imagine that this is a very painful day
for the monkey. And I'm going to get the monkey. It's behind here and I hope that you would pay
some respect to Robert. His name is Robert,
and it may take me a minute. (class murmurs) (class cheers and laughs) LEWIN:
Here is Robert. (laughter continues) I thought... (class applauds) LEWIN:
I thought it was appropriate
to change for the occasion. I don't go on monkey hunts
too often but when I do it,
I'd like to do it in style. (class laughs) Here is Robert, and we're going
to put Robert up here. Robert has in his head
a metal plate... so that when we activate
the electromagnet that we can stick him on there and when we take the current
off, then Robert will fall. So this is the activation
of the electromagnet. So here we go. I can see Robert is nervous. (class laughs) And you can't blame him. This is not the greatest day
of his life. (class laughs) Oh, by the way,
I want you to know we are not cruel here. He's wearing
a bulletproof vest. (class laughs) Oh, boy, I can... I can feel
him shaking all over his body. He's very nervous. Robert, don't let go yet! Oh, let me show you. It's important that you know that we have done
everything we can to aim this gun as accurately
as we can at Robert. Robert, don't let go yet. We've got to first cock the gun. Hold it, now, hold it, Robert! (class laughs) This happens always with Robert. (laughter) (Lewin admonishes Robert
in whispers) (class laughs) Okay, he just promised me
that he will not let go again. When I cock the gun-- if I can
find the golf ball, it's here-- then the electric circuit
takes over and now the current
will be disconnected when the gun is fired. Even I... even I'm nervous. I admit it, you know,
this is a terrible thing to do. Terrible thing to do. (sighs) You're ready? Three, two, one, zero. (gun clicks) (class exclaims) (loud echo reverberates) LEWIN:
Poor monkey. See you Friday.
