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Horzonia road) passe ; direclty beneath going 60 rri / hr A hol-Ai baboon 140 #abov tound when & mulorcycle (trnveling sbaight line than mntr of charige of he d...

Question

Horzonia road) passe ; direclty beneath going 60 rri / hr A hol-Ai baboon 140 #abov tound when & mulorcycle (trnveling sbaight line than mntr of charige of he dislarce between Ihe motorcycle ard the balloon Cecono later? (08 A /a) Iftho hillann Heoe venicAlb ric 0l ura Iha vertical disdrica fomn the balloan the rond arizbc the drlanice belaeen the Letx be the hotvraual dlstance hom the balloon thu molaicyclo molarcyclo and the ballocn Write an equation relaling * and 7

horzonia road) passe ; direclty beneath going 60 rri / hr A hol-Ai baboon 140 #abov tound when & mulorcycle (trnveling sbaight line than mntr of charige of he dislarce between Ihe motorcycle ard the balloon Cecono later? (08 A /a) Iftho hillann Heoe venicAlb ric 0l ura Iha vertical disdrica fomn the balloan the rond arizbc the drlanice belaeen the Letx be the hotvraual dlstance hom the balloon thu molaicyclo molarcyclo and the ballocn Write an equation relaling * and 7



Answers

Height of a Balloon A hot-air balloon is floating above a
straight road. To estimate their height above the ground, the
balloonists simultaneously measure the angle of depression to
two consecutive mileposts on the road on the same side of
the balloon. The angles on the road on the same side of
22. How high is the balloon?

Alright in this problem, we have a motorcyclist. He's on a nice little trip here, and we know he is traveling at a rate off 40 MPH. Okay, The hot air balloon is 100 and 50 feet in the air. Okay. The motorcyclist passes directly underneath the balloon and continues in a straight line. All right, So what we want to know is how fast the distance between the two is changing after 10 seconds has passed. Okay, so we know. Let's just call this M. We know d m d t. We didn't convert that in defeat per second, 58.67 feet per second. Hey. And we're told that the balloon is still rising. Yes, We're gonna call this be. The balloon is still rising out of rates of 10 feet per second. All right, so let's think about this a little bit. Looks like to me that we have a nice right triangle going on here. We have a right triangle. Okay, so we're gonna start with our tiger and there, um, have m squared. Plus B squared is equal to D squared. All right, so let's go ahead and take the derivative of the whole entire equation. Because nothing is stationary, everything is moving. We don't have a constant so we can't sub that anything in you So we're gonna have to em d m e t plus to be d b d t equals two d d d d t Okay, so what I'm gonna do here is I'm just gonna cross out all my choose. I'm just gonna divide the entire equation by two and get rid of those. All right, so at this point, I can start substituting and some things in. All right, So m and would be the measurement from when the motorcycle was directly under the balloon until 10 seconds later. Well, if the motorcycles traveling at 58.67 feet per second 10 seconds later would be 586.7 feet further down the road. Cmdty, we know is 58.67 now be we know it started at 150 feet. When the motorcycle passed under it, it was 100 50 feet in the air. It travels 10 feet higher every second. So after 10 seconds, it would have traveled another 1000 feet higher. Which means it is now at an elevation of 250 feet. We know DBT. We're told his 10 de. Okay, We're gonna have to calculate D. We're gonna use the Pythagorean theorem again. We know M is 586.7 were a square that and add it to 250 square. And that's going to give us. He squared. So on the left hand side here that calculates out 2406 716.89 And then we're gonna take the square root of that, and that'll give us 637.74. All right, so we're gonna have 6 37.74 and then not is gonna be times d d d t All right, so we're gonna multiply together here. 5 86.7 times 58.67 is going to give us 34,421 0.689 and then we're gonna multiply 250 times 10. That's gonna give us 2500 and then we still have 637.74 d D DT adding these two things Together we have 36 9 21.6 89 equals 6 37.74 and then we're going to divide out that 6 37.74 and we're left with a D d d T is equal to 57.9 feet per second.

Okay, So in this problem we have a hot air balloon that is traveling off the ground when somebody takes a pellet and fires it at the hot air balloon. Now the hot air balloon has a velocity of seven meters per second, and we're told it's a constant velocity, meaning there's no acceleration. Um, now the pellet is starting from the ground, and it starts with a velocity of 30 meters per second. Um, and the pellet is a free falling object as it moves towards the balloon, so its acceleration is based on the planet. It is on the negative 9.8 meters per second squared of the earth. Now this problem is asking us at what two points were that will the pellet and the balloon be at the same height? And what height is that? So there's going to be two moments in time where the balloon and the pellet occupy the same vertical height. Um, the distance that both of them travels is related, though a little bit different. The balloon will travel some distance acts. The pellet has to travel that same distance. If it's going to catch the balloon. Also, it needs to cover the 12 meters up to where the balloon is right now. So we get X and X plus 12 as our displacement values. Okay, with that, we're good to start our kin O Matics here. Now, you'll notice neither of those problems makes use of the final velocity, which means equation wise. We're going to be working with the equation that does not use final velocity that X equals V, not tea plus 1/2 a T squared. Um, the balloon is the easy one to do this for, because it does not have any acceleration. So when I plug in my numbers, we have X equals the initial velocity times T so X equal 70 and that whole 1/2 a T squared part that just become zero because the acceleration is zero. So here is our first equation for this problem. Now related the distance to the time the pellet has a somewhat more complicated equation. So when we plug that into the same equation, were placed acts with X plus 12 equals. The initial velocity multiplied by time, plus 1/2 a T squared. So negative 4.9 t squared. Um, I'm gonna subtract 12 to get everything on the same side here. And this problem asked us for two moments in time. So one thing you should be aware of that we're heading into because we're looking for two possible answers. And because we have our equation that has both time and time squared in it, we are almost certainly headed towards the quadratic equation, so we can expect to be using that to find our two moments in time. Now, at this point, I have X equals 30 T minus 4.9 T squared, minus 12. I'm going to use the substitution principle to plug in my first equation into my second equation. And we have 70 equals 30 T minus 4.9 T squared, minus 12. And at this point, I'm gonna organize it so it looks more like a quadratic equation here, um, zero equals negative 4.9 T squared plus 23 t minus 12. And for this problem for any quadratic, really, I like to make sure my a value is as simple as possible, so I'm going to divide everything by negative 4.9 to make my a value just equal to one and that gives us t squared minus 4.6 90 plus 2.45 And that equation, when we solve it, will tell us what to moments in time, these pellets will be at the exact same height. I'm gonna tuck this over here and let's break out the quadratic equation. So hopefully I'll remember that the opposite of B plus or minus radical B squared minus for a C all over two times a um So when I plug in my values, here we have my time is equal to the opposite of B. So 4.69 plus or minus the radical of 4.69 squared minus four time my A value of one times my see value of 2.45 divided by two times I, and having a equal to one always makes quadratic equation much easier to work with. So when we plug in the complicated, complicated part into our calculator, we get 4.69 plus or minus 3.49 divided by two. We'll divide both terms by two. We get 2.345 plus or minus 1.745 which gives us our two times of 4.9 four 0.6. So when the pellet is on its way up, it crosses paths with the balloon at 0.6 seconds into the problem. Um, when the pellet is on its way down, it crosses past with it at 4.9 seconds. Um, so gonna move this aside here from one more second? Because our end goal is actually figure out the final height that the pallets go to. So that's going to be that X value, Um, that we are had in our list of known values. So I'm going to plug it into the balloons equation, the one that told us that X equals seven t. That's this one over here, and that gives us our two exes of 28.6 44.2. Now we're looking for how high up the pellet got. And if you look the displacement of the palate, we need to take that X value and add 12 to it. So my final answer is the pellet will be at the same height as the balloon at 40.6 meters or 16.2 meters. And those were the two heights where the pellet has caught up to the balloon

For this problem, we have a hot air balloon rising At a velocity of seven m/s. At the instant of the hot air balloon reaches 12 m above the ground. Yeah, A gun on the ground fires a projectile going at 30 m/s. The gravitational acceleration negative 9.81 m/s squared. There are two values at which the hot air balloon and the projectile are at the same height, were to find both of them 1st. Real fine tea when they are at the same right fine T. We use the canned Mac equation X equals V, not times T plus one half a T squared plus X. Not for the bloom acceleration equals zero as it is constant velocity. Thus x equals 70 plus 12 time. When the gun fires for the gun We have x equals 30 T Plus 1/2 times negative 9.81 times d squared. And two equations We get 70 plus 12 equals 30 plus one half times negative 9.81 times t squared simplifying re range we get zero equals negative 4.905 times t squared Plus 23 T minus 12. Now we use the quadratic equation. T equals negative B plus or minus square root of B squared minus four A. C. Divided by two a. So plugging in the numbers, we get t equals negative 23 plus or minus the square root of 23 squared -4 times -4.95 times negative 12, Divided by two times negative 4.95. I used to calculate we find two values of T At 0.598 seconds And 4.9 seconds. I'll find the height that correlates to these two times for the balloon. Because is the simpler equation Which is x equals 70 plus 12. Well, and put the time of 0.598 seconds. So we get x equals seven times 0.598 plus 12, Which gives us a height of 16.19 m. We'll do the same thing for the time equal at 4.9 seconds, Which is seven times 4 nine. What's 12? Which equals 40.63 m.

So we've got a hot air balloon with a velocity okay upward of seven yeah meters per second. At a height of 12 m. There's a When it's a height of 12 m then there's a gun firing from the ground at an initial speed of 30 meters per second. So We're trying to find the two places where they have the same altitude at the same time. Okay. Well the altitude of the balloon is going to equal v. No it's not it's going to equal Y0 plus V. T. The altitude of the bullet. Oh my goodness. I guess all right. G for gun uh is going to be zero plus that. Um It's just gonna be O. V. Initial T minus one half G. T. Squared. Okay. But we want the heights to be the same. So although we also want the times to be the same but I think it's going to be easier to just set the heights to the same. Why zero plus V. T. Equals V. Sub gt minus one half G. T squared. Okay zero equals negative Y zero plus. Uh And then I'm gonna have V. G minus V. Just uh simplifying this some more honest one half G. T squared. So now I'm gonna graph this and look for the Y intercepts. Um Why Subzero is 12 V. Is seven V. Sub G is 30. Okay negative Y zero plus E. Sub G minus V. He minus one half G. T squared. Um I need to write Y equals I'm putting this in the days most and it gives me two times. Looking at the X intercepts. The times are um 0.598 seconds And 4.091 seconds. Now I just have to convert these times into positions. But um we can just use this. So why equals why Subzero whoops. Um Plus V. Yeah times T point. So that's going to be point 598 And then for the 2nd 1, same thing, why equals no plus V. I'm T. Which is 4.091. Okay, that's going to give me mm 16 0.2 meters and 40.6. Mhm. Yeah, meters. So again, I just took these times and I substituted into there and that gave me these answers. Uh huh.


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