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Calculus: Solving Vertical Motion Problems


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About this Lesson

  • Type: Video Tutorial
  • Length: 11:53
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 128 MB
  • Posted: 06/26/2009

This lesson is part of the following series:

Calculus (279 lessons, $198.00)
Calculus: Applications of Integration (10 lessons, $16.83)
Calculus: Motion (3 lessons, $5.94)

Taught by Professor Edward Burger, this lesson comes from a comprehensive Calculus course. This course and others are available from Thinkwell, Inc. The full course can be found at The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hopital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, "Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas" and of the textbook "The Heart of Mathematics: An Invitation to Effective Thinking". He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The "Journal of Number Theory" and "American Mathematical Monthly". His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

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Applications of Integral Calculus
Solving Vertical Motion Problems Page [1 of 2]
I thought we'd try to find just a couple more of these motion examples where you can really see sort of the anti-derivative power at work. In particular, let's take a look at this vertical motion issue again, whereby taking the anti-derivative of acceleration, that gave us velocity, and by taking the anti-derivative of velocity, we actually got position. So with two quick anti-derivatives, we immediately see that with vertical motion, where the only force acting on it is the force due to gravity, we see that VFt equals -16t^2 plus the initial velocity times t, plus the initial position. So that was the formula we actually derived just by doing all that integration.
And now what I'd like for us to do is to take a look at some fun examples. So the first example is one that maybe is hitting even a little too close to home. You know, you've been working really, really hard. You've been trying calculus, you've been doing it, and of course you drag out the textbook and you do the calculus stuff and you're doing all these things, and you just get frustrated and you just want to pack it in. So what you do is you take your book, which costs you a fortune, and you bring it over, and you say, "The heck with it. I'm just going to throw it away." But you just don't want to toss it away because someone might find it and see your name inside of it and return it. You want this to go and never see it again, so you take it and you actually go to a big old pit or a well and you take it and you drop it in and you listen. You can hear the screaming of the thing. In fact, I'll do that right now and you can hear exactly what it would sound like. Listen closely. Here we go. [splash] Did you hear that? Well, that dropped for 5 seconds, so that well was pretty deep. And as you walk home now sort of unburdened and feeling like a weight has lifted off your shoulders since you no longer have to carry around that 15 million pound calculus book, you start to wonder to yourself, "Gee whiz, if it took five whole seconds for that book to finally hit bottom, I wonder how long that pit was?" Well, the next thing you know you're doing calculus again. Isn't that amazing? You can't seem to avoid it. It's part of your life now.
Anyway, let's see if we can answer that question. So if we just took the book and dropped it and listened closely and heard that it landed five seconds later, the question is "How far did the book fall?" Namely, how deep was the pit? So how would you think about this problem? Well, I'll tell you how one could think about it, at least at first glance. At first glance one could think about it like this--we have a cross-sectional view that looks like this. Here's the ground, here's the pit, and then here's the ground. This is a cross-sectional view. So you're here and you're holding your book over, and in fact, in here are thousands of calculus books from former students.
Anyway, you toss the book down, it streams down and then plunks right down. That took five seconds to hear that. And the question now is how long is this? Okay. How could you think about this issue? Well, one way to think about this would be to first of all notice that we can figure out some of these things. First of all, the initial velocity we know, because all the student did was just take the book and let it go. The student wasn't sufficiently angered to actually take the book and shove it down there. He just took it and let it go. So the initial velocity in this case turns out to actually be zero.
Now what about the position? Well, you see, if we think about this level here as sort of the zero level--ground zero--that's when things hit--ground zero--the target. In fact, you know what? Now that I think about it, I think I may even have in my massive prop collection here, a target--ground zero. Can you see that? Can I have the bigger targets please? So, in fact, ground zero is down here, so let's call that level--there go you--this is where you want to hit. You want that calculus book to hit right there. So here's ground zero, and so we're going to call this level zero. And so, in fact, our initial height is going to be y^0. So that, in fact, is the height that we're looking for-- y^0. So if we insert all this information in, this particular example, we see that y(t) equals -16t^2, and this now is 0 times t so it drops out, and I have +y^0, where y^0 represents this height, the height from ground zero up to the level that you dropped the book at.
Well now, how do I find it? That's what I'm in search of. Well, what do I know? Well, I know that it took five seconds to get down there. So what do I know about the position after five seconds? Well, after five seconds I'm at ground zero. So this must be equal to zero. So if I plug in 5 in for t, I know this has to equal zero. So I plug that in, I see that -16 x 5^2 which is 25, that plus y^0 will equal zero. I plug in 5 here. Well, what does that equal? Well, what's 16 x 25? Well, 16 x 25, of course, we all know equals 400. So I see -400 + y^2 = 0 and so y^0 = 400 feet.
So as you walk back you discover that, in fact, you tossed your book 400 feet into this pit before it hit its target. That's pretty neat. All you had to use was this basic fact. Great. Well, now I want to take a look at one last example. So one last example using the fact that y(t), vertical motion, is equal to -16t^2 plus the initial velocity times t plus y^0. Okay, and now here is the question that I want us to think about. And the question is the following: I want to get into better shape, so what I want to do is I want to take this particular ball--it's not a baseball, as you can tell. It's sort of multi-colored--I think it's one of these thing for carpal tunnel syndrome or stress or something. What I want to do is I want to hurl this thing up in the air, and what I want to have happen is I want to be able to throw it so that it gets to its highest point of 25 feet. So I want to throw this thing so that when it gets to the top it's 25 feet up. The question is how fast do I have to throw the thing up in order for this thing to hit a max of 25 feet?
Well, what do I know? If I'm starting at ground level, which let's suppose we are, that means that my initial location is zero, so in this case that is zero. So in this problem I know that since I'm starting on the ground, y(t) is -16t^2 plus my initial velocity, which is the thing I'm trying to figure out, and what do I know? I know that I want it to hit a maximum height at 25 feet. When does something get to its maximum height? Well, when the velocity is zero. When it no longer moves up and before it moves down. It goes up, stops, and comes down. So when is the velocity zero? I need the velocity function. The velocity function is just the derivative of y, and so that's going to be -32t + v^0, the initial velocity, that's a constant, and I want to find out when that's zero, so I set this equal to zero, and if I solve this for t, what do I see? I see that t equals... Well, -v^0, all divided by -32, or in other words, v^0 divided by 32 seconds. So that's the time required for my ball here to get all the way up to the top.
And now the question is, "Okay, I want that top to be 25 feet." So what do I do? Well, I take this time, plug it back into position, and ask when is that position equal to 25? So let's do that. So what I want is we want y of this particular time to equal 25 feet. So when I plug in I see -16 times... And I have then v^0 squared divided by 32^2 plus... and then I have v^0 times v^0 over 32, so that's v^0 squared. I have two v^0 there divided by 32, and that's it. So this is zero. And so now I want that particular location to be at 25 feet. So now all I have to do is solve this for v^0. The first thing I think I would do is try to get a common denominator here, so I think I'll multiply top and bottom of this fraction by 32. I put a 32 here and 32 here, then I'll have a 32 squared on the bottom, and what do I have. I see 32 - 16, so that's just 16 over 32^2 times v^0 squared equals 25. And so if I solve for v^0 squared I see that v^0 squared equals 32^2 divided by 16 times 25. And I take square roots of both sides. Now, which one do I use, the positive or negative square root? Well, I'm going to be throwing this thing up in the air and that's a positive direction, so I only need to worry about the positive square root. Take this square root and see that v^0 would be... Well, the square root of 32^2 is 32. The square root of 16 is 4. The square root of 25 is 5, so this just equals 8 x 5 which equals 40. So I have to hurl this thing up at a rate of 40 feet per second in order for me to attain a maximum height of 25 feet. Neat.
The point is that just using this fact, which we discovered through anti-differentiating, we can now really uncover a complete, total plethora of examples in vertical motion. Up next what I want to do is return to the issue of anti-derivatives and look at more exotic functions, and begin to develop a new method to allow us to find the anti-derivative of functions that are even more interesting than the ones we've already seen. Come back when you're ready, and I'll see you there.

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