34) An arrow is shot upward at an Initial velocity of 64 feet per second:The height of the arrow is given by the function below where h IS the height in feet and t ...

ARCHERY An arrow is shot upward with a velocity of 64 feet per second. Ignoring the height of the archer, how long after the arrow is released does it hit the ground? Use the formula $h(t)=v_{0} t-16 t^{2}$ where $h(t)$ is the height of an object in feet, $v_{0}$ is the object's initial velocity in feet per second, and
$t$ is the time in seconds.

Or this function nature that tells us the height. We want to determine the maximum height of our ever as well as how long it takes to reach the ground. So let's first go ahead and forgot the max height. Now this is a quadratic, and it's shaped like this is going to be facing downward. So the maximum value is going to occur at its Vertex. So to find that we can let t equal to negative B over to a where A, B and C are the coefficients of squared regular term and just are constant. And then we can plug and so would have negative 1 60/2 times second of 16 which is negative. 32. So those negatives cancel, and then we just have 1 60 over 32 which is five. So now this tells us the time it takes and to find our height, we're going to take five and plug it into our core tracts of B H of five equal to negative 16 times five squared plus 1 60 times five plus eight. So we would get 25 times 16 so it's gonna be negative or 100 plus 1 60 times five, which is 800 then plus eight. Now that's going to give us 400 plus eight, which is, or +08 And the units with this. I'm just double check should be in feet now to find the time it'll take to reach the ground. But that, really is asking is when is a chav t equal to zero? So let's go ahead and set this equal to zero. So that's going to be negative. 16 t squared Lost 160 t lost. No notice that we can divide all this by negative eight. Let's do that really quickly. So we have negative eight. And I just want to do this toe help simplify the numbers a little bit so we don't have to work with as large numbers, especially of us to t squared minus 20 T minus one. Let me just double check to negative 20. Negative one. Yeah, that's good. Now we can go ahead and plug this into the quadratic formula to help us soul. So that's gonna be T is equal to negative B plus or minus the square root of B squared minus for a C all over to it. And again we didn't need to divide by negative eight. We could have just used original one. But at least now we can plug in smaller about use those book everything good. So negative B is going to be 20. What survives to square root of negative 20 squared, which is or 100 minus for a C. So it's four times two times negative one, which is going to be positive. Eight all over two times two, which is going to give us four Now, actually, let me screw this up a little bit to give us some more room now. Adding those together we'd get 20 plus surmised, the square root of or eight and we can actually factor out. I want to say a four from that. So one so eats divided by four is one or two, so this can be re route as two times the square root of 102 And then does it look like we can actually simplify that any further? So we'll end up with two solutions. 20 plus or minus two. Rule one No. Two over four, but noticed that if we check the negative one is going to give us a negative value, so we would go ahead and get rid of that negative since a negative time in this case doesn't make any sense. And we could just go ahead and approximate what this value is, and it looks like it would be around. So 20 plus two times 102 divided by four. So somewhere around 10.0, or so this is the exact time with the approximate time over there.

It's a furthest physics problem. We're trying to figure out which two times, Ah, that the era was going to hit 32 feet above the ground or B 32. And that's gonna happen twice because what goes up must come down. So one point goes up to 32 goes past 32 feet, then comes back down and will be at a rate of, ah, height of 32 again when it comes down. So we're going to plug in 32 is equal to 48 t minus 60 t squared. And so we're gonna clean that up and get this to be, um, in the proper order and everything, so we're gonna have zero is equal. And so we gotta change the sun. Uh, so we're gonna have negative 16 t squared plus 40 it t and then minus 32. So what we're going to dio is we're going to go ahead and divide everything by a negative 16. Or in fact, we could just do the quadratic formula and see what we get sought. So I'm gonna do so. Negative sixteen's are a 48 R B and C is negative. 32 so we're gonna have the opposite of be so negative. 48 four time from his Mabel Negative 48 plus or minus the square root of 48 cleared minus four times negative 16 times negative. 32 all over two times a year of 16 which is a negative 32. So we need to clean up that square root stuff quite a bit and it's gonna be a negative times, a negative times, a negative, meaning it's going to keep it being negatives. They're gonna have 48 squared, which is 23 04 minus the four time 16 times the 32. And that's going to give us 2 56 as our total um, under the square root. So it's gonna be 1948 plus or minus that screwed of 56 all over Negative 32. Well, a square root of 2 56 IHS 16. So that is going to equal negative 48 plus or minus 16 over to get 30 to so And in this case, we're gonna have both of them become positive answers, and these answers will actually develop because a negative out of my negative will become positive. So we're actually gonna have to valid answers, as we stated before. So we're actually have negative 48 plus 16 over negative 30 to we're gonna have negative 48 my esque scene over negative 30 to so negative or April 16 is negative. 32 over negative three to gets us a positive one second. So that's one or answers one skins, one of our answers and the negative 48 plus negative 16 is negative. 60. We're over. Get third. That will give us a positive of two seconds. Well, so it's going to go up, and at one second it's gonna be 32 feet above the ground is gonna keep going up. But then, with gravity and physics, what goes up must come down. So when it comes back down two seconds later, it's also at 32 feet above the year. But then it just keeps going down until it hits the ground again.

Alright. Looking at this problem here, we've got the height of something being launched in the air. And then we were also given Intergraph now part a tells us to use their graph to estimate how long this object is in the air. Well, we start off here at time T equals zero notice over here. That point right there which is a time T equals for its height, is also zero there. Because if you look at this the Y axis here, that represents your height and this was whenever hits the ground again and so we expect it to be in the air about four seconds, about four seconds now be tells us to do it. Algebraic lee. So it's going to hit the ground again when it's hide a zero. And so it actually solve this rhythm, putting in zero for age of tea and then solve this for T's. We have zero equals negative 16 t squared plus 60 40 and then we have a quadratic to solve. And so we're going to solve this by factoring to pull out a negative 16 t That leaves you with T minus four on the inside. And so using the zero product property that's tells us native 16 t zero or T minus 4 to 0 if name search team T A zero that tells us tea is zero. It just means it's on the ground whenever you launch it. All right, so that doesn't tell you how long it's actually in the air. This other part, though, does your tea is four foot 64 seconds before it hits the ground again. And so that's what we should have on beat. I don't see says to use their draft s maiden win reaches its maximum height, so I compared the graph. What time does it look like it is whenever it reaches this peak up here? Well, it's like it's at about two seconds. So we expected to reach its peak at about two seconds. Yes. We're going to say about two seconds now. When do we actually want to find an algebraic Lee? So we have a job t his name of 16 t squared loss of 64 teams. Now, if you think about this here, this is a problem, and so probably is gonna regions. But a just point at its vertex. And so here. We just need to find the X coordinate or, in this case, the T coordinate of the Vertex. So the T coordinate of the vortex performing was negative. Be over to a So that's negative. 64/2 times Negative. 16 a 64. Don't be negative. 16 during this is his nature of 64. Over. Negative 32. It would be positive too. So it'll take it two seconds before it reaches its maximum height.

So this question here is telling us that we have shot. So it's just trying to get you to understand how this, um what this what these equations mean in a really world situation and in this case, physics. So the question is asking us if we shoot a projectile straight into the air and the initial velocity is 64 feet per second. They tell us that the height is going to be represented by this equation here, and I've just rearranged it to be where the higher exponents t is first, just because that tends to be how we write it. And so what that means is, if I'm standing on the ground and I shoot a projectile straight into the air, it's gonna go something like this. And so what does this look like? The same parabolas that we've been using and looking at all along. So what they want to know is, what is the height and feet and t and and seconds does it take to get to the highest point? And so what they want to know is, how high is this? And also how long and seconds did it take to get to that height. And so all we're looking for, then is that for tax point and that Vertex is a maximum. And so what we do is just use the same equations that we've been using the entire chapter. So what we're looking for is first, where is how long does it take for it to get to this height? And that's gonna be our access a cemetery. And so we can just use that same equation. Where is the negative b over to a And so here are B is 64. So this is a negative 64 over, two times in negative 16. And so this is gonna be negative. 64 over a negative 32. So that gives me two. And remember that tea was in seconds and ah, the height was in feet. And so now we are gonna use that to and plug it back in to find out what the height is at two seconds. And so when I do that, I have that the height is equal to a negative 16 times two squared plus 64 times two. And so I have remembered to do the order of operations in the correct order. And so first, I'm going to do exponents. So a two times a two time, the two are a two squared is four. And so this gives me a negative. 64 plus 1 28 And so this is 64 feet. So our height at the maximum point is 64 feet, and it takes us two seconds to get there.