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Quadratic word problem: ball

Sal solves a word problem about a ball being shot in the air. The equation for the height of the ball as a function of time is quadratic. Created by Sal Khan and Monterey Institute for Technology and Education.

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Video transcript

A ball is shot into the air from the edge of a building, 50 feet above the ground. Its initial velocity is 20 feet per second. The equation h-- and I'm guessing h is for height-- is equal to negative 16t squared plus 20t plus 50 can be used to model the height of the ball after t seconds. And I think in this problem they just want us to accept this formula, although we do derive formulas like this and show why it works for this type of problem in the Khan Academy physics playlist. But for here, we'll just go with the flow on this example. So they give us the equation that can be used to model the height of the ball after t seconds, and then say about how long does it take for the ball to hit the ground. So if this is the height, the ground is when the height is equal to 0. So hitting the ground means-- this literally means that h is equal to 0. So we need to figure out at which times does h equal 0. So we're really solving the equation 0 is equal to negative 16t squared plus 20t plus 50. And if you want to simplify this a little bit-- let's see, everything here is divisible at least by 2. And let's divide everything by negative 2, just so that we can get rid of this negative leading coefficient. So you divide the left hand side by negative 2, you still get a 0. Negative 16 divided by negative 2 is 8. So 8t squared. 20 divided by negative 2 is negative 10. Minus 10t. 50 divided by negative 2 is minus 25. And so we have 8t squared minus 10t minus 25 is equal to 0. Or if you're comfortable with this on the left hand side, we can put on the left hand side. We could just say this is equal to 0. And now we solve. And we could complete this square here, or we can just apply the quadratic formula, which is derived from completing the square. And we have this in standard form. We know that this is our a. This right over here is our b. And this over here is our c. And the quadratic formula tells us that the roots-- and in this case, it's in terms of the variable t-- are going to be equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. So if we apply it, we get t is equal to negative b. b is negative 10. So negative negative 10 is going to be positive 10. Plus or minus the square root of negative 10 squared. Well, that's just positive 100, minus 4 times a, which is 8, times c, which is negative 25. And all of that over 2a. a Is 8. So 2 times 8 is 16. And this over here, we have a-- let's see if we can simplify this a little bit. The negative sign, negative times a negative, these are going to be positive. 4 times 25 is 100, times 8 is 800. So all that simplifies to 800. And we have 100 plus 800 under the radical sign. So this is equal to 10 plus or minus the square root of 900, all of that over 16. And this is equal to 10 plus or minus-- square root of 900 is 30-- over 16. And so we get time is equal to 10 plus 30 over 16, is 40 over 16, which is the same thing if we divide the numerator and denominator by 4 to simplify it as 10 over-- or actually even better, divide it by 8-- that's 5 over 2. So that's one solution, if we add the 30. If we subtract the 30, we'd get 10 minus 30. Or t is equal to 10 minus 30, which is negative 20 over 16. Divide the numerator and the denominator by 4, you get negative 5 over 4. Now, we have to remember, we're trying to find a time. And so a time, at least in this problem that we're dealing with, we should only think about positive times. We want to figure out how much time has taken-- how long does it take for the ball to hit the ground? We don't want to go back in time. So we don't want our negative answer right here. So we only want to think about our positive answer. And so this tells us that the only root that should work is 5/2. And we assume that this is in seconds. So this is in 5/2 seconds. I wouldn't worry too much about the physics here. I think they really just want us to apply the quadratic formula to this modeling situation. The physics, we go into a lot more depth and give you the conceptual understanding on our physics playlist. But let's verify that we definitely are at a height of 0 at 5/2 seconds, or t is equal to 5/2. This expression right over here does give us h is equal to 0. So we have-- let's try it out. We have negative 16 times 5/2 squared plus 20 times 5/2 plus 50. This needs to be equal to 0. So this is negative 16 times 25/4 plus-- let's see, if we divide 20 by 2, we get 10. If we divide 2 by 2, we get 1. So 10 times 5 is going to be 50. Plus 50. This needs to be equal to 0. Negative 16 divided by 4 is negative 4. 4 divided by 4 is 1. So you have negative 4 times 25, which is 100. Plus 50-- oh, sorry. Negative 4 times 25 is negative 100. Plus 50, plus 50 again is equal to 0. And so we have negative 100 plus another 100. Well, that's definitely going to be equal to 0. We get 0 equals 0. And it all checks out. We hit the ground after 5/2 seconds. Or another way to think about it is 2.5 seconds. t is equal to 5/2 seconds, or 2.5 seconds.