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. E $2.44$ ). After the sandbag is released, it is in free fall. Compute the position and velocity of the sandbag at $0.250$ s and $1.00$ s after its release.

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. E $2.44$ ). After the sandbag is released, it is in free fall. Compute the position and velocity of the sandbag at $0.250$ s and $1.00$ s after its release.

Diagram showing a balloonist releasing a sandbag at 40.0 m height with velocity 5.00 m/s.

Accepted Answer

everyone in this problem, We have a Montgolfier balloonist moving up with a uniform speed of 10 m per second. They're going to drop a bag of stones from the balloon when the balloon is 50 m above the ground, And we are asked to find the speed and position of that bag at .5 seconds after its release. And we're told that after the bags dropped it's under freefall. All right, so let's think about this. When the bag is dropped from the balloon, it's gonna have the same initial speed as the balloon. Okay, so when that bag is first dropped, it's going to actually be moving upwards at 10m/s, just like the balloon. So let's take up to be positive. Okay, so that means that our initial velocity in the Y direction is going to be 10 m/s. Now V. F. Well, we don't know the final velocity. That's what we want to find out. That's one of those things we want to find out is the speed, the acceleration while we're under free fall. So we have the acceleration due to gravity, it acts downward, so it's gonna be negative. They have negative 9.8 m per second squared. Okay, t we're told to consider after 0.5 seconds, we have t is equal to 0.5 seconds. Okay? And then we have delta Y and delta. Why we don't know either, but we want to know the position of the bag. Okay, so this delta Y is something that we also want to find out. Alright, so thinking about our um equations, Okay, we have three of the unknowns. We want to find the others. So we can go ahead and use our equations. Let's start by finding VF. Okay, so when we're finding VF we have V. Not A and T. We want to find the F. So we choose the equation that doesn't include delta Y. We're gonna have v. F. is equal to v naught plus 80. So V F is equal to 10 m per second plus the acceleration negative 9. m per second squared times the time, 0.5 seconds. Alright, so simplifying, we get 10 m/s minus 4.9 m per second squared times second. This is also going to be meters per second and then we can subtract and we get that V F is equal to 5.1 m per second. Okay, and so that is the speed of the bag after 0. seconds. Okay, so we figured out the first part of the question. Now we need to find the position. Okay, so we've done VF check, we've done VF. Now let's go and do delta y. So same thing we have V not we have a we have t okay, we want to find delta y. So let's use the equation that doesn't include VF delta y. And really in this case you could use any equation because now we found VF So we do have all four quantities. You could use any equation. Let's just go ahead with the three quantities we found initially delta Y is equal to v naught. E plus one half a t squared. Okay, so delta Y. It's going to equal the not the initial velocity which is m per second times the time. 0.5 seconds plus one half the acceleration minus 9.8 m per second squared times a times squared, 0.5 seconds squared. Alright, so all of these seconds we see are going to cancel. We have second divided by second and then we have second squared divided by second square. We're going to be left with five m -1.225 m. Which is going to give us a delta Y. And let me move. So we have a bit more room. A Delta Y of 3.775 m. Okay. Alright, so Delta Y is 3.775 m at 0.5 seconds. Now, let's think about this. We've chosen up to be the positive Y direction. So far, Delta Y is positive. Okay, that means that we've traveled upwards, which makes sense because our initial velocity right, is the same as that balloon traveling upwards. Okay, so our change and why is upwards and we want to note the position relative to the ground. Okay, so the height above the ground? Well this is going to be equal to the initial height when the bag was dropped plus whatever our delta wise, how much we've changed after the 0.5 seconds. Okay. Well, when the bag was dropped, we were at 50 m And after 0. seconds we've gone 3.775 m. Okay, so this is going to give us 53. m above ground. Okay. Alright. So have our final speed and now we have our final position. Okay. So the answer is going to be D after 0.5 seconds, we're going 5.1 m per 2nd and 53.8 m above the ground. Thanks everyone for watching. I hope this video helped see you in the next one.

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Key Concepts

Free Fall

Kinematic Equations

Initial Conditions