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Ch 02: Motion Along a Straight Line

Chapter 2, Problem 2

A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground (Fig. E2.44). After the sandbag is released, it is in free fall. (d) What is the greatest height above the ground that the sandbag reaches?

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everyone in this problem, we have a Zeppelin balloonist, it's moving up with a uniform speed of 10 m per second, it's going to drop a bag of small rocks at the instant when the balloon is m above the ground. Okay? And we're asked to find the maximum height above the ground of that bag of rocks. Now, we're told after the bag is dropped its under free fall. Okay? So let's think about this. When the bag is dropped from the Zeppelin, it's going to have the same speed as the Zeppelin balloon. Okay, so the Zeppelin balloon is moving up at a speed of 10 m per second. Okay, so let's take up to be positive. This means that the initial speed of the bag in the Y direction, It's going to be 10 m/s. Alright, So if this bag is initially going upwards, it is going to attain some height above where it started. Okay, So let's say it gets to a maximum height here. Okay, well, we're gonna call this Delta. Why? Hey, and how do we find this Delta? Y. Well, let's write out the other information, we know we know that Vienna is 10 m per second. What about V. F. The final velocity? Well, when the bag reaches the maximum that we're looking for, we know that it's going to come to a stop momentarily. Okay, when it gets to its maximum it will momentarily stop before coming back down. And so the final velocity we're looking for is going to be 0m/s, that's gonna be when that bag stops right at the top. The acceleration, we know we're told it's under freefall. So we're gonna have the acceleration due to gravity. This is going to be negative 9.8 m per second squared. Ok. It's negative because the acceleration is acting downwards delta Y. This is what we're trying to find. Okay, And then we have time. We don't have information about time and we don't want to find anything out about the time so we can ignore that. Okay, We need three variables. We have the three things that we know. We want to find delta Y. So let's go ahead and use are you am equation. Okay, that doesn't have tea in it. And that's gonna be the following. V F squared is equal to V not squared plus two. A delta wife substituting in the information. We know V F is zero. So the left hand side is just gonna be zero. We have v naught squared 10 squared Plus two times a negative. 9.8 times delta What? And our units here, 10 is meters per second. A is meters per second squared so we can move everything with the delta wide to the left hand side. We're going to get two times 9.8. Okay, Because we're moving it over. We add so the negative is going to go away and let's add the units in here. So this is meters per second squared. Okay, Delta Y. And on the right hand side we have 10 squared which is 100 in our unit here, we had meters per second and we squared it. So now we're gonna have meters squared per second squared. Okay, We can divide by two times 9.8 we're going to get delta Y. Is equal to 5.1 in our unit. We have meters squared per second squared, divided by meters per second squared. So we're just gonna be left with the unit of meter, which is exactly what we want when we're talking about distance. Alright, So we have delta Y. So we have that what that change in height of the bag. Okay, But the question is asking the max height above the ground. We're talking about the max height above the ground. We'll call it h Okay, well this is gonna be the initial height plus that change In height that change and why that we found because the bullet or the bag is at some initial height when it gets dropped, it rises that 5.1 m that we found. Okay, so its total height Again starts at 50 m when it's dropped, it rises 5.1 m before it starts to fall downwards. Okay, so this is going to give us a height. Maximum height of 55.1 m. That's going to correspond with answer a that's it for this one. I hope this video helped. Thanks everyone for watching
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Textbook Question
A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground (Fig. E2.44). After the sandbag is released, it is in free fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release.

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