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(~I5 Points]DETAILSAUFEXC4 10.4.041_MY NOTESASK YOUR TEACHERPRACTICE ANOTHERKarenthmowincoranaeher brotner Saul, "hostundinoDulconytheir home The height_feet, ...

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(~I5 Points]DETAILSAUFEXC4 10.4.041_MY NOTESASK YOUR TEACHERPRACTICE ANOTHERKarenthmowincoranaeher brotner Saul, "hostundinoDulconytheir home The height_feet, tne orande adove che dround seconds after It is throvn given byh(t)16t22.118.Saul $ outstretched arms feet above the ground_ Nes - Orange willreacn maximmm height of 19tnc Orangc @cihinn cnouanthat he can catch it?The orange 4reacrmaximum helght . not enough information given answc the question;TherThe OfangeMatnmmm height = feet: The

(~I5 Points] DETAILS AUFEXC4 10.4.041_ MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Karen thmowinc oranae her brotner Saul, "ho stundino Dulcony their home The height_ feet, tne orande adove che dround seconds after It is throvn given by h(t) 16t2 2.118. Saul $ outstretched arms feet above the ground_ Nes - Orange willreacn maximmm height of 19 tnc Orangc @ci hinn cnouan that he can catch it? The orange 4reacr maximum helght . not enough information given answc the question; Ther The Ofange Matnmmm height = feet: The orange wIll reach maxlmum helaht of 38 feet. Neod Help? Feea eetealn



Answers

Review Interactive LearningWare $2.2$ at www.wiley.com/ college/cutnell as an aid in solving this problem. A hot air balloon is rising straight up at a constant speed of $7.0 \mathrm{~m} / \mathrm{s}$. When the balloon is $12.0 \mathrm{~m}$ above the ground, a gun fires a pellet straight up from ground level with an initial speed of $30.0 \mathrm{~m} / \mathrm{s}$. Along the paths of the balloon and the pellet, there are two places where each of them has the same altitude at the same time. How far above ground are these places?

So in this problem, we're going to use some Kine Matic equations to figure out the figure out the Parametric equations of of e b ball that is kicked with respect your time and so and so we have this. This position of X is equal to the initial velocity times the coast I have the angle of times t plus the original the original starting position X and then so for y t. Since since our Y component has grabbed the acting on it So we have this negative 1/2 a cheese square buzz be not sign A t, which is the initial velocity plan is a sine of the angle attempts t plus the initial initial Why by position and so so if we use this well, we can say all right, What we know the you know that the X component should be X is equal to X is equal to We have this initial velocity of 1 40 times. He co sign angle of 45 45 times t and and this problem we can assume that that our initial exposition is euro. We have this X is 1 40 coastline 45 times tea. Then we have. Why is equal to Well, we have this negative 1/2 of the of acceleration, right? So in in acceleration in terms of feet is negative is 32 sides negative 1/2 times 32 which is negative. 16 the negative 16 t squared. She squared plus again, you in your velocity 1 40 This time we have sign of the angles or sign 45 times two plus our initial life commission, That is. He danced in the u zero. It was. Look at this graph. We look at this graph. Well, we can figure out when the exposition is. We can figure out what the exporting it is right here. Yeah, if we figure out where. Well, this is why component is zero, and we can figure out where the Y component is. Zero, If we re say Okay, well, let's factor out 80 the same tea we have. Negative. 16 plus 1 40 sign. 45. This people, this is the negative. 16 t negative. 16 to you. Plus 1 47 45 This should equal zero. And so Well, we already know what we have. These two points T is equal to zero. Then we have to solve for the inside negative 16 TV is so making 16 t plus 1 40 signed by is equal to zero. When we do that, we do 1 40 45 divided by 60 we get six point. He is equal to 6.2. You see what you 6.2? And if we plug in 6.2 x what we get that our exposition after the ball hits the ground. So remember, the ball goes, uh, and then it's the ground. So when X is at this yes, 6.211 time is six point to This should give us her our range. So how far how far did this ball go? This is This is what we're trying to figure out right now. How far do this ball coach? So let's play and much. 6.2 13 to find how far the ball went. So get 6.2 times. One for you. Co signed 45. So this X is 613 point white age eat. Let's late with this. This is seconds. This is also seconds. And so we found that so So if a T is equal to six point to the ball hits the ground again. Well, if this is a parabola, then we know that at some point along here, half of the distance half of the time because it takes half the time to go all the way up to its maximum height and then half the time to go back down. So well, if we divide 6.2 by two, get 3.1. So what? We know we hit our maximum height. AT T is equal to 3.1 seconds. And so if we saw a full if we saw for why, what we get is negative. 16.1. Thus, 1 40 five 3.1. What this gives us is 153. So So our Max Max height max height is 153. Why one It's this is distance, family distance travel. And so so this. Yes. Yeah, is Max I travel. And so let's see. Let's see if we're missing anything. Well, we found we found Max height. That's a check. We found the distance traveled. We even found we even found our parametric equations. right here. So we have this permission equation Eggs of X with respect to T and I will respect t so we have effectively solved our problem. We have solved department, remember? Remember the formulas we used this these are These are dramatic formulas for position of a of A ball and it uses, uses gravity, uses initial velocity and use this These two equations are very, very important problem. So I've labeled labeled what each each very mo means. Remember, this is this is the trajectory of the ball. And we use reasoning to figure out that that if we can solve for for our endpoint here when the ball hits the ground in other words, were y zero where are wife of one of the ball is your way solved for that? And our X component is when is at the time where the ball hits the ground. So this is our distance travel. I'm liking in 6 22 unplugged in 3.1 because it's half of 61 too. So somewhere between those have half the time it takes for the ball to go up and back down. So we want to figure out where those maximum up and this maximum up our maximum wise A maximum height which we figured out to be 1 53 So we've checked all of our all of our questions and we are done.

So in this problem, we're being asked to estimate the height of some objects from stone statue. So here's my statue. We're trying to figure out the height. So to do this he stood 20 feet away from the statue. So compared to the base of the statue, which would be a right angle, he's 20 feet away. And we were told that the angle of elevation from where he standing to the top of the statue is 70 degrees, which means this bottom angle in our triangles, 70 degrees. So again, we want to find the height of the statue and we're gonna round to the nearest foot. Well, compared to the ankle were given in the triangle. We're trying to find the opposite side ever given the adjacent side. So we're going to use the tangent function to do this. So we're gonna have the tangent of 70. Degrees is equal to the opposite side, which is X divided by the adjacent side, which is 20. And now we just have to solve for X. So to do this, we're gonna multiply both sides of our equation by 20 and now we just have to punch this into the calculator. Now make sure when you do this that your calculator is in degree Moon and we were told to round to the nearest foot, meaning to the nearest whole number. So what you'll find is that the statue is approximately 55 feet.

Question Number 54 question of the E. The equation will be s ft equal minus 16 T squared plus 90 t question number be projectile height function given in shell velocity 19, 3, 90 and initial height. Zoo so t equals minus 90. Over to multiply by minus 16. Equal minus 19 over minus 32. Equal 45 over, then equal 2.8125 So is equal. Minus 16 multiplied. Boy T point, uh, 125 square plus 90 multiplied by two points 8125 Equal one, um, 126.56 seconds. Question number is T makes me hide for quadratic formula A T squared plus B t is when t equal minus B over to eight. Use that value to draw mine as at that time. So minus 16 square plus 90 t is bigger than 120 then minus 16. He squared plus 90 t minus 120 is bigger than zero, so G equal minus 90 plus or minus square root 90 square minus four multiplied by minus 60 multiplied by minus 120 over. To multiply by my multiplied by minus six. Texting so the equal 2.17 or T equals 3.4 five. Second, solve quadratic and quality. 40. Subtract 120 quadratic formula. So question number MM is equal minus 16. Q. Squared plus 90 t so it's equal. T multiplied by minus 16 T plus 19 and then zero equal minus 60 t plus 2019. So 16 t equal 90 and t equal 5.625 seconds. Factor quadratic equation into different form. Factor out T rocket heads ground when as equal zero Ignoring t equals zero because that's pre launch. We only care that s equals zero when minus 16 T plus 90 equals zero sold for tea.

In this time we're throwing a rack up. And so it would look something like this. Yeah, when you throw something up, it always has a parabolic shape. So this is time. This is our height. And so we want to find a formula that will represent the situation. And so the formula we can use the why is equal to one half. I am gravity time squared. And then we add to that foolish initially last eight times time. And so in this case, we're told her initial velocity is equal to 90. And so that's why my love will become Why is equal to G here? Um, when we're doing feet per second is approximately negative. 32. So one half times negative. 32. It's about negative. 16 last nine million times time. Okay, so you actually want to find out the max height And so to find out the max height, we can actually recognize something here. Is that this travel? It's metric. And so the time it takes to reach the end, um, it will reach the max height halfway in between the starting time in the end time. So, actually, if we find the end time, we can cut that in half and give us the max hype. So let's find the ending time where that will be when our problem is equal to zero. Because the height would go down to a height of zero, which is the end. So we can factor a t here. So we get negative 16 t plus 90 0. And so this gives us two solutions equals zero would give us here, or if this part equals zero. So that happens. The negative 16 plus 90 put it zero. Which means that t she put 90 16, which is approximately 5.6 to 5. And so this is actually part D. Look, um, parte de asked us to find the time it takes to reach the bottom, so that's 5.6 to 5 seconds. And so the to reach the max I, which is what Pope is asking, boy is just half of that. So, probie, we do one half times 5.6 to 5, which gives us approximately 2.8125 seconds. And so we want to know the height at that point. So we dio height is equal to negative. 16 times are approximately 2.8 square. Yeah, plus 90 times 2.8. And this gives us approximately 126.56 z That party on. Finally, we need part C, which is how long is it above 120 ft. So the way we can do that as we set our height, which is a negative 16 t squared, that's just 90 80. And we want to highlight to be larger than or equal to 120. So you can do that by solving this equation. Negative. 16 2 squared waas 90 t minus 120 a zero. And so we can solve this by using the quadratic equation. We're a here is make them 16. It is 90 and she is negative 1. 20 and that would give us a solution. The two solutions, we get 2.17 seconds and 3.45 seconds. So between 2.17 and three point 45 it will be above 120 ft. So that is part C. And we already did part the earlier That was 5.6


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