###### Alissa Fong

MA, Stanford University
Teaching in the San Francisco Bay Area

Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts

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# Applications of Quadratic Equations - Problem 2

Alissa Fong
###### Alissa Fong

MA, Stanford University
Teaching in the San Francisco Bay Area

Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts

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The problem we’re about to do is about a diver jumping off a diving board. But you’re going to want to use these same ideas any time you see your homework or test problem about something flying through the air, and I promise you you’re going to have some. It’s going to be like a base ball or a soccer ball or a firework, a canon shooting somebody or whatever, it's going to involve something flying through the air.

So let’s go ahead and read it carefully. A diver starts on a platform 50 feet above the pool. Assume he’s starting up with velocity v, is 6 ft per second. Use the equation h(t) equals -16t² plus vt plus s; where s is the initial height, to find the number of seconds, t before he hits the pool.

Okay if I were a student in a high school Algebra and I was looking at this problem I’d probably skip it too. So don’t be too hard on yourself if this is really intimidating. I’m a Math teacher and I’m kind of intimidated by all this language. Look at this equation we have like, 1, 2, we have 3 variables plus it’s in function notation; there’s a lot of intimidating things so we have to go through step by step.

First thing I like to do is draw a picture. I’m not a very good artist so you’re going to have to bare with me here. Here’s the diving board, here’s our little guy about to jump, a little Speedo on him. Okay so he’s about to jump and then he’s going to come down and hit the pool. And we know he is 50 ft above the pool when he starts. That’s what they told me; he’s 50ft above the pool. So this amount right here is 50; from the diving board down to the pool. And then in my picture you can see how it’s an upside down parabola right, when he flies through the air he’s going to go up a little bit and then he’s going to come down. I’m trying to find how long it takes for him to complete this path. So it makes sense that my equation starts with a negative coefficient on my t² term. Negative remember is going to means it's going to be an upside down parabola. So that makes sense.

Okay, let’s talk about all these other stuff that’s going on. V stands for initial upward velocity. They told us that starting upward velocity v. That’s like when you jump off a diving board you kind of bend your knees and go. So you start out, you don’t start out from like zero velocity, you start with a little bit of speed you’re not just like accelerating like a car or something. So this guy starts with an initial velocity and it looks kind of scary when they told us that v is 6 ft per second. So wherever I see v in that equation I’m just going to write a six instead. I’ll rewrite that in a second, I’m just trying to help myself remember that v is 6. They also tell me something about s, s is the initial height. Okay so we know how high this guy starts. He starts 50 ft above the pool. So when I rewrite my equation, instead of the letter s right there, I’m going to use 50. So let me try to clean this up for you so it doesn’t look so ugly. The height of the guy in terms of time is going to be -16 times time in seconds squared plus 6 times time plus 50. This I can deal with. This only has one variable and that makes me a lot more comfortable.

Okay, now the last and most important thing in this problem that most students are going to miss is trying to figure out what height we’re looking for. What we’re trying to find out is we want him to hit the pool. When he hits the pool, he’s going to be zero feet high right? He started 50ft high; when he hits the pool he’s going to be zero feet high. So I’m going to substitute for h, height I’m going to call that zero ft high. Now I have a really straight forward problem using a quadratic equation that I can solve using a number of different methods. So it’s up to you if you want to use quadratic formula, completing the square, factoring, taking square roots, probably you wouldn’t want to take square roots because, oops that should be a 6 right there. You would take square roots if you didn’t have a b term but we do have a b term.

So I personally, first thing I noticed is that 2 is a common factor into all of these. I’m going to divide all of these by 2 so that I have some smaller numbers to deal with. You don’t have to do that. You’ll still get the same answer. Just for me it makes it less intimidating because now my numbers are at least a little bit smaller.

Okay the next thing I like to try to do is factoring. I’m a pretty good factor except for my leading coefficient is 8. So that means if I were to try to factor I don’t know if that would be 1t and 8t or maybe 2 and 4 also 25 has a couple different factor pairs so I’m deciding not I don’t want to use factoring. It looks like it's going to be too much guessing and checking for me. The neat thing about solving quadratic equations is there’s a couple of methods that always work. Completing the square always works and the quadratic formula always works. I personally I’m going to use the quadratic formula; it’s totally up to you if you want to or not, you might choose a different method. This is the quadratic formula that you guys have memorized, b² minus 4ac all over 2a. Don’t worry I’m not going to sing the song today.

What I’m going to do is go through and use my coefficients, substitute them into this formula to find my x value. By the way, I use the letter x here which is kind of a mistake, our equation technically has the independent variable t so I’m going to erase that and make that guy a t because we are solving for t. Okay so my b value was 3 plus or minus square root of 3² take away 4 times a times c all divided by 2 times -8. Okay, I’m going to back up a little bit so I have a little more space. I’m going to also jump ahead because I’ve already done this on my calculator, I know you guys can do this simplification just be really careful with the negative signs. In there I’ll have the square root of 809 divided by -16. Okay so this plus minus tells me I have two different answers; I’m going to have one where I’ll do -3 plus square root of 809 is 28.44. I’m also going to have another answer that comes from -3 minus 28.44. These both need to be simplified further. When I do this, I end up getting -1.59. That’s my one answer for t. my other answer for t when I simplify this is 1.96.