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About this Lesson
- Type: Video Tutorial
- Length: 14:17
- Media: Video/mp4
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- Posted: 07/01/2009
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This lesson was selected from a broader, comprehensive course, Physics I. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/physics. The full course covers kinematics, dynamics, energy, momentum, the physics of extended objects, gravity, fluids, relativity, oscillatory motion, waves, and more. The course features two renowned professors: Steven Pollock, an associate professor of Physics at he University of Colorado at Boulder and Ephraim Fischbach, a professor of physics at Purdue University.
Steven Pollock earned a Bachelor of Science in physics from the Massachusetts Institute of Technology and a Ph.D. from Stanford University. Prof. Pollock wears two research hats: he studies theoretical nuclear physics, and does physics education research. Currently, his research activities focus on questions of replication and sustainability of reformed teaching techniques in (very) large introductory courses. He received an Alfred P. Sloan Research Fellowship in 1994 and a Boulder Faculty Assembly (CU campus-wide) Teaching Excellence Award in 1998. He is the author of two Teaching Company video courses: “Particle Physics for Non-Physicists: a Tour of the Microcosmos” and “The Great Ideas of Classical Physics”. Prof. Pollock regularly gives public presentations in which he brings physics alive at conferences, seminars, colloquia, and for community audiences.
Ephraim Fischbach earned a B.A. in physics from Columbia University and a Ph.D. from the University of Pennsylvania. In Thinkwell Physics I, he delivers the "Physics in Action" video lectures and demonstrates numerous laboratory techniques and real-world applications. As part of his mission to encourage an interest in physics wherever he goes, Prof. Fischbach coordinates Physics on the Road, an Outreach/Funfest program. He is the author or coauthor of more than 180 publications including a recent book, “The Search for Non-Newtonian Gravity”, and was made a Fellow of the American Physical Society in 2001. He also serves as a referee for a number of journals including “Physical Review” and “Physical Review Letters”.
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Projectile motion is free-fall in two dimensions. It means that an object is acting under the influence of gravity. It's traveling through space. And we can understand its motion very simply. The acceleration of gravity is down and it's constant. So the equations that we have to work with are the one-dimensional equations of motion in the x direction and in the y direction, separately. That's the beautiful thing about two-dimensional motion, is you can always think about it as what's going on in the x world and what's going on in the y world? You can treat those separately. You use the same set of equations with the different constants that are appropriate. And at the very end of the problem, if you know x final and y final, you know where the object is. You know its motion. So the key to projectile motion is to separate the problem into x and y.
Let me show you a picture of projectile motion. Here's an object, which was launched sideways. Looks like it had a slightly downward initial velocity, and it's following some kind of a projectile motion. And let's just think for a second about what is the velocity as a function of time; not the position as a function of time, but the velocity. So look at the equations of motion. First let's focus on the x equations of motion. Remember that if gravity is the only force, if acceleration is straight down, then the acceleration in the x direction is zero, so . It never changes. You can see that with these little green components of the velocity vector. This is not the velocity, and neither is that. It's the sum of those two. This is the x component. That's the y component. So the velocity vector itself is always tangent to the curve. And you can see that its horizontal piece is never changing. It's a little counter-intuitive, but absolutely correct physics.
What about in the y direction? The y equation of motion for projectile for velocity is . Acceleration in the y direction, I'm calling down positive, is +9.8 meters per second squared. So velocity in the y direction is increasing linearly with time. If you double the time, your velocity is twice as much. The way I think about it is acceleration is change in velocity with time, 9.8 meters per second per second. So each second my downward velocity is 9.8 meters per second more in this direction than it was the second before. You can see that. It started off small. It's getting bigger and bigger.
What about the speed as a function of time? We haven't talked about that. Explicitly, speed is the magnitude of velocity. It's the length of the total vector. It would be by the Pythagorean Theorem . It's clearly getting bigger and bigger, because this component is constant and that component is getting bigger. That makes sense. A falling object is going faster and faster if it starts off nearly horizontal. And you can just see it pictorially.
Let's ask ourselves are we set up to solve the most general problem? I mean have we really finished the story in two dimensions? It's fairly impressive, but we really have. All you need are those one-dimensional equations and you can solve any problem that you want with constant acceleration, and in particular, any free-fall problem. Here's a rather generic two-dimensional projectile problem. This is ground level, and you launch something at some angle theta with some initial speed V. I could write it as a vector V, but what I'm telling you is what's the speed and what's the angle. That's all you need to know. And I'm telling you where I launched it from. I'm going to call this x, and I'm going to call y up just for arbitrariness.
I would like to ask, well there are lots of things you could want to know about this. Fundamentally, the big question is what's x as a function of time? What's y as a function of time? Once you know that, you know the motion. You can find velocity. You can find anything. It's a little bit imposing to have such a generic problem. Let me make it specific. I've got this little toy gun and I'm going to launch it. And I would like to know, if I just fire it, how far is the arrow going to go? So where is it going to land? It's going to fly through the air, hit here, and I would like to know the range. That's the word that people usually use. Range refers to how far something goes if it's traveling across level ground. We might give it the name R.
That's the puzzle. How do we find R? Of course it's one arbitrary puzzle, and I could come up with lots more. How high does it get? How long does it take? Let's just solve the first problem first. I'm asking about a horizontal distance. So here comes the usual story when you're working a projectile problem. I've got six equations, the three x equations, the three y equations. Where do I begin? Well I'm asking a question about the x world. I want to know how far it gets, so I immediately focus my attention on the x equations. And I've got three of them. And I'm asking a question about x final, basically. I could call this zero in the x world. So I really want to know x final. That's my unknown. So look at the equations that I have to work with. There's no acceleration in the x world in projectile motion. So the equation is very simple . So alas I can't solve for x final until I know the time. I do know the x component of the initial velocity, because if I've got the magnitude and the angle, it's just geometry. V[ix] is just V cosine theta. And V[iy] is just V sine theta. So I really do know the initial x and y velocities.
I feel stuck. I can't find x final until I know time. How do I find the time? Well it seems like there's really nothing I can get out of the x equations. There's nothing more I can get. I certainly can't learn about the time from the x equations. How might I learn about the time? Well I've only got one other set of equations. Let's look in the y direction. An object in the y direction is going up and back down again. Its x motion is independent of its y motion. So as it travels along this path, it's just climbing up and falling back down again. And I want to know the time when it gets back down to the ground. So what information is that? When it hits the ground, it's basically saying y final is zero. Y initial is zero, y final is zero. So I jump to the equation . I know V initial y, . That's an equation which has only one unknown, time. And so I'm all set. It's a quadratic equation, but fortunately, there's no linear term. Let me write it down. In this case I have I have --what's the acceleration in the y direction? I was calling up positive. So this is -9.8 meters per second squared. And I'm completely running out of space here, times t^2.
I want to solve this for time. How do you do that? You don't even need the quadratic equation. Sometimes you need the quadratic equation, but here I'm okay. Let me factor out time, . And I've got one more parenthesis I have to match up. This one was really kind of unnecessary. There's an equation. It's time times something equals zero. If time times something equals to zero, there's only two possibilities. Either time equals zero or something equals zero. Time equals to zero, that's a correct solution to this equation. Remember this equation was when is the object at height equals to zero, and the answer is at time equals to zero. I knew that.
But there is another answer, when this is equal to zero. And let's be careful. This is really V[iy]. , I know what that is. Let me just solve this little internal equation. If this equals to zero, I get . Formula, time equals, I know V[iy]. I know what g is. So I know how long it takes, but that wasn't the question I was after. I wanted to know the range. But I'm all set. I know the time and the range. I go back to my x formula. X final, range, equals x initial plus v initial in the x direction times time. Let me plug in the formula that I just got for time. V[iy] divided by g over 2. So there's a nice simple looking formula. It's equal to V[ix] V[iy] times 2 divided by g. And that's V cosine theta. And that's V sine theta. So I can write it like so. Let me put the 2 upstairs, because it's in the denominator of the downstairs. Here's the range formula. So if I tell you what theta is and what the initial speed is, a nice convenient formula to find the range. It's a handy formula to have, but a little word of warning, remember this is only the range if you start and end at the same height. There will be lots of problems where you start and end at different heights and then you can't use this formula.
Let me make one last little simplification here, so we can sort of stare at this formula. It's a lot of good physics already just in this first formula we've derived. Sin theta cos theta times 2, there's a trig identity. That's the same thing as sine of 2 theta. Range is equal to V^2 over g sine of 2 theta. That means that the faster you launch, the farther you're going to go. Makes sense. What about the angular dependence? As you launch at different angles, what does that tell you about how far it's going to get. If theta equals to zero, it just plows right into the ground, and you don't get anywhere. If you launch straight up, theta is 90, 2 theta is 180. The sine of 180 is 0. You launch straight up. It goes straight back down. It doesn't go forward. Remember this is the x motion. What's the optimal? The optimal is when the sine is 1. That's the biggest the sine of anything can ever be. That would be when 2 theta is 90 degrees, or theta is 45 degrees. So that's the best you can ever do. If you want this thing to fire as far as possible, pick 45 degrees.
Let me make a little comment about the real world here. I tend, like all physicists, to get into my physics world where there is no friction and life is simple. In the real world, there is some air friction. And what does that do? Well it complicates the story. You see we no longer just have an acceleration that's down. We now also have a frictional force, which makes an acceleration that's countering the motion. So the friction's acceleration is always changing in direction. It's a very difficult problem to solve. You need really a computer to solve numerically. We've got the equations. Acceleration tells me change in velocity with time, so I can figure out velocity. There's no simple formula. Basically, if you've got air friction, as you can imagine it won't get as far. We're not really going to worry about that too much. In this course, from time to time, we will investigate questions of friction coming up.
What have we done? We have solved, essentially, all two-dimensional physics problems with constant acceleration. I've only done one or two specific examples, but you've got the equations. You can solve them yourself. Any problem I give you, any question that I ask, you're all set up to solve it. What about three dimensions? Do we have to have a whole other tutorial on going to three dimensions? No, three dimensions is the same as two. It's just another set of equations. It's the same set of equations, right? The three usual equations, just put a sub z in all the variables, and you've got equations of motion in three dimensions. So really we've solved the whole story for constant acceleration.
Where are we headed? Well we know everything there is to know, basically, about motion in two and three dimensions with constant acceleration. We could still ask some interesting kinematics questions like what about objects that run around in circles? If you're running around in a circle, that's going to turn out not to be constant acceleration, but it is something that we can understand. That's one place we've got to go. And finally the big place, what's physics really heading toward, is why? We've been discussing what's the motion. I'd like to know why the objects move the way they do. We'll get there.
Investigating Motion in Two Dimensions
Understanding Projectile Motion Page [3 of 3]
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