A 124 kg balloon carrying a 22 kg basket is descending with a ...

Question

A 124 kg balloon carrying a 22 kg basket is descending with a constant downward velocity of 20.0 m/s. A 1.0 kg stone is thrown from the basket with an initial velocity of 15.0 m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. That person sees the stone hit the ground 5.00 s after it was thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 m/s. At the instant the rock hits the ground, how far is it from the basket?

Accepted Answer

Hey everyone welcome back in this problem. We have a tool bag attached to a pulley, Okay, And that's gonna be bringing that tool bag upwards. Something's going to happen, There's gonna be a spring that comes from inside the tool bag and launches itself outwards and it's going to be launched perpendicular to the path of that tool bag. Okay, so tool bags going up, it's going to be launched perpendicular early. Um after three seconds the spring is going to hit the ground, okay? And the tool bag is going to continue to rise while the spring is falling. All right, and we're being asked when it hits the ground, how far is that spring from the toolbox? So, the first thing we want to do is to draw a diagram of the system and get an idea of what's going on and what exactly we're trying to find. Alright, so, let's draw what's happening initially. So, we have our pulley, okay, And it's going to be attached to our tool bag here, This is our tool bag. It weighs five kg. It also has a toolbox within it, that is three kg. Okay, so that's the information we're given. five kg tool bag, three kg toolbox within it. Alright, and we're told that it has the upward velocity of 12 m/s. So, this pulley is pulling this up at 12 m/s. Alright, And the spring is going to come out of that bag, and I'm just gonna fall kind of like this. Okay, Alright, so the spring we're told is 500 g 500 g Spring converting that into kg. It's just 0.5 kg. Okay, we want to have consistent units for everything and its initial velocity is going to be 10 m per second. And again it's perpendicular. So the bag is going up straight. That means that perpendicular we're going to take to the right, So this is going to be 10 m per second, that the spring has its initial velocity. Okay? And then the spring is gonna fall. All right, So, after the spring falls, we have our pulley and the bag is going to have been moved up. Okay, so pretend this is the same bag. It is now 4.5 kg because it's lost that spring. So, it's lost half a kilogram because of the spring. And then we have the ground. We'll draw the ground here. We have the ground and the spring has now landed on the ground. Okay, It's been three seconds. And we want to know the distance between the spring and that bag. Okay, So we want to know this distance here, and we're gonna call that distance D Alright, well, if we consider the X. Direction of this distance, Well, that's just going to be how far the spring travel. Ok? In the X direction because the spring started with the bag, it's ending on the floor. So that's gonna be the change in X. Of the spring to the displacement in the X direction of the spring. Alright. In order to find d. We also need to know this why component. Well, we know that the spring started here. Okay, this was the springs initial launch position. Okay, so this distance from where the spring started down to where the spring finishes is going to be the change displacement, delta Y. Of the spring. Okay, And then this top distance, Well this is where the bag started, The bottom is where the bag started and it goes up to where the bag is now. So that's gonna be the displacement in the Y. Direction of the tool bag. So this total Y distance here, we can call H. And it's just going to be the sum of the change in the bags, wide displacement and the change in the springs. Why position. Okay, Alright, so we have our diagram, let's go ahead and sort this out. So let's start with this spring. Okay, so we need to find some information about the spring, Delta X. S. And delta Y. S. Okay, so these are our intermediate variables that will need and a D. Again is a distance we're looking for. Alright, so again, starting in this with the spring. Okay, let's just list out the information we know. Okay, we have the X direction and we also have the Y direction. So we don't want to forget about both directions now in the X direction. Well we know that the initial velocity is 10 m per second. It's to the right which we've taken as positive. Okay, So take to the right and up as positive. So I forgot to include that in the sketch at the beginning, so to the right and up is going to be positive. So our initial velocity in the X direction is going to be 10 m/s. The final velocity in the X direction. Well, we don't really know the acceleration in the X direction while the acceleration in the X direction once it leaves that bag, there's no external forces acting on it in the X direction. So the spring actually won't be accelerating in the extraction. So the acceleration is just gonna be zero. Okay, we know that the time is three seconds. Let me just scroll down so we have a bit more room here. We'll keep our diagram up so we can refer to it. Alright, so the time is three seconds and then delta X. Of the spring. Well, we don't know that, but that's what we're looking for. Okay, that's gonna give us this X component of the distance we're looking for. Alright, so in the X direction and let me actually erase this wide direction here first, let's work in the X direction first. All right, So, if we look at are you am equations Okay, we're looking for an equation that doesn't contain the final velocity but has the other variables. Okay, And that is going to be delta X. And in this case it's the spring, it's going to be V not T plus one half a t squared. Ok. And again in this case the acceleration is zero. So this term is gonna go to zero and our delta X. S going to be equal to the velocity in the extraction 10 m per second. Okay. Times the time which is three seconds. Okay, the unit of seconds will cancel and we're going to be left with 30 m. Okay, so delta X s is 30 m. Okay, so we can go ahead and add that on to our diagram. Okay, so this quantity here, The change in the exposition is gonna be 30 m. Alright, let's do the same for the spring in the Y direction. Okay, so again its initial velocity in the Y direction. Well initially it's traveling upwards with that bag. Okay, so when it first leaves that bag it's gonna have the same velocity as the bag. So it's actually going to be traveling at 12 m per second and it's going up. So that's gonna be positive, we don't know anything about the final velocity. Mhm. The acceleration well we have the acceleration due to gravity so that's gonna be acting downwards so in the negative Y direction, so negative 9.8 m per second squared In the time again, it's the same amount of time, three seconds and delta Y of the spring? Well that's what we're looking for. That's going to be part of this H component, part of that total Y component that we need to find our distance. D. Okay so so we're looking for delta Y. S. We don't know anything about V. F. Okay. So we're gonna choose the equation that doesn't use V. F. And in this case it's gonna be the same equation we use in the X direction. We're gonna have delta Y. S equals V. Not Y. T plus one half a T squared. And in this case we do have an acceleration so that second term isn't gonna cancel. So substituting our values, we have 12 m per second, the initial velocity times a time of three seconds plus one half the acceleration minus 9. m per second squared times times squared three seconds. All squared. Working this out, we're gonna get 36 - for whoops. And let's put the units in here, sorry. So again we have meters per second time second. The second will cancel. And we'll be left with meters and then for the right hand turn we have 44.1 and here we have meters per second squared times seconds squared. Okay, so the second squares were canceled and we'll be left with meters. So the units are going to match their the unit of distance that we want and we get minus 8.1 peters. Alright, so that's our delta Y s minus 8.1 m. Okay. And the reason for the negative is is because the spring is going downwards, the spring is falling and we've taken up to be positive. Okay, so when we're looking at h that total y distance we just want to take the absolute value. We just want the magnitude of that distance. Um not the direction. Okay, so delta Y s 8.1 m. Alright, so the last thing we need to do in order to find d while we need to find this delta Y. B quantity. Okay, so delta Y B is going to be the change in the y position of the bag. B. Alright, so let's give ourselves some more space. Alright, so when we're looking at the bag, okay, we know that its velocity Initially is 12 m/s. Okay, and it's positive because it's going up, we're told that this is a constant speed. Okay? So we're told it's a constant speed, that means the acceleration is going to be zero. We know that three seconds has passed and we want to know what the change in the y position is. What is delta Y. Of the tool box. Okay, so in this case we can ignore the final and choose the equation that doesn't use it. Okay, so it's gonna be the same as the equations we've used for the spring, we're gonna have delta why of the toolbox is going to be the initial velocity plus one half a t squared. Ok? Again, in this case the acceleration zero. So just like the X direction of the spring, that second term goes to zero. Okay. And we just have 12 m per second times three seconds. That's gonna give us 36 m. Okay? So this is our delta Y. T. That we were looking for. Alright. So if we just scroll up to our diagram for a moment, so now we have delta excess 30 m, we have delta Y. S. 8.1, we have delta Y. B. Which is 36 m. Okay, so now we can figure out d. So let's go back and we're just going to draw another diagram to use here. Alright, so this is what we kind of have above. We have This distance which is Delta X. S. which is 30 m. We have this distance and we know that the top part of this is delta Y. Of the toolbox, which we found to be 36 m. Okay. And the bottom part is that change how far that spring fell? Delta Y. S. And we're just gonna take the absolute value. Again, we're just worried about the magnitude of that distance. 8.1 m. Okay, so that means that this total length of this side is going to be 44.1 m. Okay. 36 m plus the 8.1 m. That's gonna give us 44.1 m. And what we want to find is the distance D. Here, I'm a hippopotamus. Alright, well we know how to do this. We have a right hand triangle. We can use pythagorean theorem. So we have D squared Is equal to squared. Okay, so that's the X. Component plus 44. squared. That's the Y component. Okay, We are going to get D squared is equal to 2844.41. And our units here are unit 34. Unit 44.1. That's meters. So we're gonna have meters squared because we're squaring the quantities When we take the square root, the unit will become just meters and we'll get plus or -53. m. Okay? And we're looking at a distance, we just want the positive value. We just want the magnitude of that distance. So we're gonna take positive 53. m. Okay, so that is the total distance between the bag and the spring once the spring hits the ground. Okay? So d. 53.34 m. And if we go back up to our answer choices, that is going to correspond with answer a thanks everyone for watching. I hope this video helped see you in the next one

Key Concepts

Relative Velocity

Projectile Motion

Vector Addition

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