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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?

$y_{1}=40.672 \mathrm{m}$$y_{2}=16.186 \mathrm{m}$

Physics 101 Mechanics

Chapter 2

Kinematics in One Dimension

Motion Along a Straight Line

University of Michigan - Ann Arbor

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Lectures

04:34

In physics, kinematics is the description of the motion of objects and systems in the frame of reference defined by the observer. An observer has to be specified, otherwise the term is meaningless.

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In mathematics, a position is a point in space. The concept is abstracted from physical space, in which a position is a location given by the coordinates of a point. In physics, the term is used to describe a family of quantities which describe the configuration of a physical system in a given state. The term is also used to describe the set of possible configurations of a system.

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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

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