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Applications of Quadratic Equations - Problem 2 4,153 views
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.
Okay so let’s think about what these mean. I’m going to go back over to my picture so we can think about like in the real world what we’re talking about and we can try to decide which is our correct answer. Okay so if you look in my picture where I have the guy on this diving board, we see that he reaches zero ft high at this point in the parabola. But keep in mind in Math we could also describe this side of the parabola which would technically be like time in the past. It’s like if this guy were to fall down or something using the same equation he would hit zero feet high at some time in the past that’s whey I got that weird negative answer. That tells me don’t want that weird negative answer like negative time doesn’t really make sense. I’m looking for the positive time answer for when he hits the water. So the positive answer that I got was 1.96 seconds. Make sure to label it with seconds so your teacher knows you remembered what you were doing, like how this applies to the real world and that’s the final answer. It makes sense right? If you ever jumped off a diving board, you know you’re not in the air for like 15 seconds, it’s like jump, do your flip and you’re down. It’s like 1 ½ seconds. So that makes sense. So it’s always a good idea to think about whether or not your answer makes sense in the real world.
So that was a really challenging problem especially because of all the weird variables and because I got some weird fractions and numbers I’m not familiar with but just stay with the process. You guys are good at doing the quadratic formula, you know how to do these problems you just need to be careful to pay attention to detail along the way, don’t lose any negatives and be sure that you’re doing the right order of operations to get your square root values.