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# 4: The Motion of Projectiles

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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.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

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