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
where 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 the 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
a 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,
the 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.
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.
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 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 )
Now I wonder what happens if I fire the ball at 30 degrees?
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 ready?
You are?
Three, two, one...
(ball whooshes )
(Lewin shouts )
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?
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 figures
and unhappy with another set of figures.
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, xt1, 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 time t
minus one-half gt squared.
Well, if there is no gravity, this term doesn't exist
because 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 ready?
Three, two, one, zero.
(gun clicks )
(class exclaims )
(loud echo reverberates )
LEWIN: Poor monkey.
See you Friday.