College Algebra: Solving a Projectile Problem

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Probably the most valuable and important application of the whole quadratic equation thing is gravity. You know, gravity is an amazing thing. If you take something, you can see gravity right there. There's gravity. It's amazing, sometimes loud, too. But actually when things fall or if you throw something up and then watch it come down, that path is actually dictated by a parabolic kind of curve or a parabola or actually a quadratic. These are all the same things. So, in fact, any time you throw something up and watch it come down, you're actually working with, and maybe not even knowing it, you're working with quadratics. So quadratics aren't as weird as you may think. In fact, let's take a look at an application and see a quadratic literally in action.
So what do we have here? Well, we have a disgruntled Algebra student, not good, but also fairly popular unfortunately. So we have a disgruntled Algebra student and she's so fed up with the whole course, she decides she's going to take her book out to a field and just give it a good hard kick; the kick that it deserves. She gives it a kick, goes up, comes down. The question is how long was it in the air for? How long was that book airborne?
Well, actually we know the path or we know, at least, the information about the path of the book when it left the ground and when it came back down and it's given by again, since we have gravity, it's given by a quadratic formula. So let me show it to you. If I let s represent the position off the ground the book is, then s is actually going to be a function of time, depending what time it is. And it turns out that it's -16t^2 + 88t + 2 where t represents seconds. So what this means is the following. If you give me how many seconds after she kicked the book, I could plug in that time for t and this will tell me how high the book actually got, where the book is located. So, for example, let's try a very simple example to make sure we really understand what this means. What about when we start, what about the moment she kicks the book? That's a times 0 we're starting. Where is s located? Where is the book located at times 0? Well, if I plug in a 0 for t, then these terms just drop out and I see 2. So that means that it's two feet above the ground? Does that make sense? Well, sure, because, you see, she was sort of holding the book and apparently she was holding the book two feet off the ground before she gave it that kick, you see? So that's the location of the book when she started and then one second later you can figure out how high the book went and the book sort of went up, up, up, up, and then came down, down, down, down. And I want to know how long was the book in the air.
Well, let's think about that. When does the book finally land? What would its distance from the ground be? Well, if you think about it, if the book is going to go up, up, up, up, up, and then down, down, down, down, boom, the book will finish its journey the moment it touches the ground; which means its distance from the ground, which is s, would be 0. It would land. So what I have to do is find out what time makes this thing equal to 0. That's the moment that it landed.
So what I do is I set this equal to 0 and solve. So I have -16t^2 + 88t + 2 = 0. And you notice one thing I could do immediately, happily, is factor out a common factor of 2 everywhere and then divide it out which sort of reduces the numbers and, of course, that makes me so happy because I can't deal with big numbers. So I'd have a -8t^2 + 44t + 1 = 0 and now my mission is to solve that. So, of course, what do we do first? Well, we're going to try to factor. Well, this is great. Life is wonderful. It's going to so well. I'm so happy. Are you happy? I hope you're happy because I'm doing great.
All right, now I've got find some way of putting together, breaking up a -8t into 2 pieces. For example, it could be -8t and t. It could be -4t and 2t and so on and so forth. Let's just try, for example, -4t and t. This tells me that these signs are going to be what? Well, I've got to think about this for a second. If I have a plus sign here then these signs are actually going to be the same and now what's the same going to be? Well, I've got to be actually a little bit careful here because I see this negative way out in front. So I should be a little bit careful because that negative is actually going to have an affect on things. So but the same sign we're going to have here. So let's see. Well, what you get if it was 1. I could put like a 1 and a 1, for example. Now how could I make that combination of 44? This seems like an obscenely high number. Obscene given the 1 and the 8 here. In fact, I've got to tell you this is not looking too good. Remember how wonderful life was just like three seconds ago? It's amazing how things turn. This can not be factored. So much for life being wonderful. All right, so what are we going to have to do? We're going to have to bring ye ol' quadratic formula. So let's use the quadratic formula and solve this baby.
What do we see? I see t equals--now what's the quadratic formula? There it is. So what is it? -b which is this coefficient all divided by 2a. We have to insert that and remember the players. The role of a is going to be played by -8. The role of b is going to be 44 and finally tonight in the role of c we're going to have a modest little person 1. Let's plug in and see what we get; -b so that's going to be -44 which is 44^2 - 4ac. So that's 4 x a x c. Well that gives me a -8 and I'm subtracting so I have a +4 x 8. So that's all under the square root, all divided by 2a and 2a is going to be -16 in this problem.
What a big old mess. So what do we see here? Well, we could clean this up a little teeny bit, I guess. Not much though. Plus or minus the square root. Let's figure out what that number is. So now I'm going to actually use a calculator, hopefully successfully we'll see. Time will tell. I'll take 44 and square it and I'll add to it 4 x 8 and see what that equals. That's going to equal the number 1968. Something that is almost in my memory. That shows you how old--I'm so old. I don't even want to get into that right now. It's not good. So there's that number and what do I do with that? Well, first of all what is this number right here? You can try that on the calculator again. We could bring up the calculator and try that number and what we'll see is the actually equals around 44.36 something. So now I have to figure out which of these two answers, if any of them, make sense. t represents the time, right? Like how long the book has been in the air. So let's see what answers we could possibly have here. This negative sign down there is something that's maybe concerning. This minus sign here is concerning. Which one of these do you suppose we should pick? Suppose I picked the positive sign. Let's look at just that one, the positive sign. Well, this number is actually a little bit bigger than 44 so that's a positive bigger than 44 and then I subtract 44. So the top is actually going to be positive, but then I'm dividing by -16 so the net total here is going to be a negative number. So that means t equals something negative. That means I've gone backwards in time. Well, that's not going to happen. She can't kick up the book, watch it come up and down and it turns out the book landed before she kicked the book. That's not going to happen. All right. So that means that, in fact, the right sign must be the negative sign. And the positive sign must be an extraneous root. So what happens? I have -44 and then I subtract off a 44.36 something and then if you do that and divide by 16, you can compute this on a calculator just to get a sense of what this is, you'll see 5.52 something seconds.
So the book was airborne for 5.52 seconds so about five and a half seconds roughly speaking so that's a pretty good kick. So took the book, kicked it up and so forth and, of course, the book lands. As you can imagine the book sort of looking in not too good of condition here which brings me to my bonus question, the final bonus question. Will the present condition of this book have any affect on the amount that student will receive when she sells it back to the bookstore? And, of course, you know the answer is no. No matter how good the condition is she'll get probably a buck fifty for this $80 book. What a great deal and what a great problem, too.
I'll see you at the next problem.
Equations and Inequalities
Word Problems with Quadratics - Applications
Solving a Projectile Problem Page [2 of 2]