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Ball launched into the sky at 272 feet per second from the roof of a skyscraper 1,344 feet tall The equation for the ball's height hat time tseconds is h =-162...

Question

Ball launched into the sky at 272 feet per second from the roof of a skyscraper 1,344 feet tall The equation for the ball's height hat time tseconds is h =-162 272t + 1344 When will the ball strike the ground?

ball launched into the sky at 272 feet per second from the roof of a skyscraper 1,344 feet tall The equation for the ball's height hat time tseconds is h =-162 272t + 1344 When will the ball strike the ground?



Answers

A ball is thrown upward from the top of a building. Its height $h,$ in feet, after $t$ seconds is given by the equation $h=-16 t^{2}+96 t+200 .$ How long will it take for the ball to be $88 \mathrm{ft}$ above the ground?

Okay, so the lowest type we want is 39 feet. Um and then our equation is negative. 16 T square, Our initial velocity is zero because we dropped it. Ah, but our initial height is gonna be 120 feet, and the larger position we're looking for is gonna be 56 feet. It's It's strictly between. So I'm gonna use the less than and greater than and not have it having equality. I guess the first thing we do is subtract 120 from all my side. So we have 39 minus 1 20 so it's gonna be negative. 81 is less than negative. 16 T square. They were gonna do 56 minus 1 20 which is negative. 64. Next thing I would do is divide everything about negative 16. So it's gonna flip our order, and we're gonna have four is less than T square is less than 81/16 and then we're only looking for a positive time. So, like when we take the square root were all they look at the positive square roots. We don't have to do another change. So take the square root of all sides leapt to his lesson. T is less than nine number four, which is like 2.25 Okay, so an interval notation We could say that the time that it's in those distances would be from two seconds to 2.25 seconds. Okay, Thank you very much.

This question it sees fine. How long does it take for the ball to reach the maximum height? And every time they're asking us to find the maximum height or the time taken to reach the maximum might as if we have a probably here upside down. And we need to get the eggs value off the verdicts because the eggs value is the time taken here, the time the value of tea to get the maximum hide each year. So we need to find the exit off the verdicts, which is negative. B over to A You're in this case, it's gonna be negative. 24 over two times negative 4.9 That's gonna be 2.448 approximately equal to point for five seconds. That's the time taken. That's the X value off the verdicts or the time taken to reach the maximum

We're given the height function for a baseball that starts out five foot above the ground when it's hit and were asked to figure how long it's in the air. Well, we know that when the height gets back down to zero, it's on the ground again. We weren't told a specific method to solve this one with. So I am going to use the quadratic formula we're trying to figure when the highest zero we have our equation. We know the quadratic formula in this case, T equals negative B plus or minus the square root of B squared, minus four a c all over twice. A. So if we fill those in, we'll have t equals negative. 76 plus or minus the square root of 76 squared minus four times negative 16 times five all over negative 32. We have negative 76 plus or minus. Inside the square root 76 squared is 5776 four times negative 16 times five would be negative, 320. But since we're subtracting that, we will put minus negative 3 20 or plus positive. So now we have t equals negative 76 plus or minus the square root of 6096 over negative 32. The square of seven or 36,096 is approximately 78 0.7 So I'm gonna fill that in now. We're gonna find two solutions here, but one of the answers is a negative number. It can't be a negative time, so we only want to look at the positive time. The positive time is going to come when we subtract the 78 0.7 So I have negative 76 spine A 78.7. That's negative. 1 54 0.7 Divided by 32. Show me divided by a negative 32 which, if I fill that in 1 54.7 negative divided by 32 negative. The time I get is four point 815 seconds

Mhm. Hello. So this problem today is pretty cool. Um It's a physics space problem. Um What could be really useful in real life because this is actually a real application of physics. So the first thing I like to do is draw a picture, so that's just what I'm gonna do here. So let us think about the problem a little better. So we know we have a little person maybe, maybe not a person who does not want to draw that's okay and he's going to throw a ball up, it's gonna come back down and hit the ground at some point in time. All right, here's the ground, here's your little ball. This is just the trick trajectory of that ball. Um Also going to draw up this line here just so you know, kind of see if this is like a parabola what that would look like. Okay so we are asked two questions in this on first part A is asking us um what at what to you what time in seconds? Um Well the ball hit the ground so the ball hit the ground we know S. Equals zero. I says the displacement where it's at relative to the ground. The second question we ask is what is t the time in seconds when the ball hits this point? So it's going to come up and it's going to go back down there asking when is it going to be at the same level of the building? The same point as which it was thrown from. So when asked equals I mean it was thrown from a height of 96. Um Yeah so with that let's just get started. We are going to start by just substituting. So for party. Yeah. You know the S equals zero source. A zero equals 96. Close A. T t minus 16 T squared. Um And this is a little different form than we're used to but it's still zero equals C. Plus the eggs clause X squared. So just backwards. But the same thing that we're used to doing so to solve this we're just gonna plug into the quadratic formula. This reminder that says negative B. Plus or minus square root of B squared minus four. A. C. Over to A. Is going to equal zero. And this is where the Y. Value of equal R equals X. Or the Y. Value equals zero for a problem. So using this formula I'm going to say X equals we should get to values of X whether they're real or not is something we could deal with later. So minus B. It's minus 80 plus or minus what square root minus 80 quantity squared minus four times a minus 16 times C. Which is 96. Okay this all is over two times. They just negative 16. Yeah So when we saw out for this scroll up a little bit here uh huh. When we yeah we saw we have minus 80 plus or minus the square root. Um 12,544 over 1932. And then we can just simplify that again. The squared of this. Just plug it into your calculator is negative 80 plus or minus 112. Bye bye. Negative 32. So negative 80 plus 112 divided by 32. It's going to give us negative one. You have um 32 or 32 then our other answer it's negative 80 minus 112 or 32 gives us six. So if you think about it we know this is a time in seconds. They're both one second. This is six seconds Um Sense the ball up here. Um Is it being thrown up and coming back down negative one second. Doesn't make any sense as to when it be at that um When it hit the ground it shouldn't hit the ground backwards. Even if we continue the problem that might make sense right? But that's not going to happen. So you know it's got to be the six seconds. So six seconds is her first answer answer to party. So for part B we're being asked when it's going to hit that equivalent point. So we know the S equals 96. So once again I just got a substitute in 96 equals 96 plus A. T. T minus 16 T squared. So 96 minus 96 0 equals E. T minus 16 t squared. It's going to add this 16 T squared to the other side here, 16 t squared equals 80 T. We can divide by one T. On both sides. We have 16 T equals 80. And if we divide 80 by the 16 we have T equals five. So at T equals five seconds. Should do that envelope blue A. T equals five seconds. You know, that's when it's going to hit the equivalence point. It will be the same distance from the ground as where it was thrown initially.


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