00:04
Okay, so this problem, we're given the mass of the balloon, which we can call uppercase m here.
00:11
So the mass of balloon is going to be 124 kilograms.
00:15
We're also given the mass of the basket, so it's going to be 22 kilograms.
00:22
And then we're given that the balloon is descending with a constant downward velocity of 20 .0 meters per second.
00:30
So that means the y component of the initial velocity is going to be negative 20.
00:35
20 .0 meters per second.
00:38
And we're also given that a 1 .0 kilogram stone is thrown from the basket with initial velocity of 15 .0 meters per second.
00:49
And it's being thrown perpendicular to the path of the descending balloon as measured relative to the person at rest in the basket.
00:58
Okay.
01:00
So we can call the mass of the rock m0.
01:03
So it's going to be 1 .0 kilograms.
01:06
And we know the rock is being thrown perpendicular to the descending balloon.
01:12
So it's perpendicular.
01:14
That means it's going to be in the negative x direction here.
01:19
So we're going to have the initial velocity in the x direction as negative 15 .0 meters per second.
01:29
It's perpendicular to the descent in the y direction.
01:35
Okay.
01:40
And it also says that the person sees the stone hit the ground 5 .00 seconds after it was thrown.
01:47
So we'll take time t to be five seconds.
01:52
And we can assume that the balloon continues its down descent with the same constant speed of 20 .0 meters per second.
01:59
So it's going to just continue descending downward at that rate.
02:05
Okay.
02:07
And for part a, it's asking how high is the balloon when the rock is thrown? so for that, we can use kinematics.
02:34
So we can use this kinematic equation for part a.
03:58
And now, so we know the initial height is going to be y0.
04:03
The final height's going to be y.
04:08
And then we have the initial velocity in the y direction is b0y.
04:14
And we have t.
04:15
And we have ay, which is the acceleration of the y direction.
04:19
So we're taking the y component of the command equation.
04:27
So we're being asked, how high is the balloon when the rock is thrown? so that means we do know the initial height y0.
04:37
So we could take the final height y to b0.
04:41
And that's when the balloon's going to hit the ground.
04:55
That's when the rock's going to hit the ground.
04:57
Yeah.
04:58
So this is actually the trajectory of the rock.
05:01
So the rock is going to start when the balloon is at the initial height, y, 0, and the rock's going to hit the ground when y is zero, because the ground level is a height of zero.
05:20
So this we can take y0, and we know v0y, because we're given that.
05:29
We know the time t, given that in the problem also.
05:33
And a y is going to be the acceleration of gravity okay because gravity is acting downward in the y direction so a y is going to be negative g it's going to have a negative one -half g t squared and now i want to solve for y zero because that's the initial height so this height right here is the height that the rock is being thrown at so i want to solve for y zero i just change the signs of these other quantities here okay so i know everything so i can plug in the value for v0y, t, and g.
07:27
So it's going to be the initial height of the rock of 223 meters.
07:42
So that's light the rock is being thrown at.
07:52
And then part b is asking, how high is the balloon when the rock hits the ground? okay.
07:58
So now we're going to estimate the height of the balloon once the rock is hitting the ground.
09:03
So we want to find the height of the balloon once the rock hits the ground.
09:11
Okay.
09:13
So we know the balloon.
09:14
Isn't accelerating so it can do so be it's not accelerating because it just at a constant velocity going downward on 20 meters per second so that means there's no acceleration term here for the balloon so we have we're using kinematics equation like we did for part a but the last term is zero because there's no acceleration for the for the balloon so that gives us this then so now we could solve for the height um why be zero here or sorry, yb.
10:07
This we basically want to change in height that the balloon is gone.
10:38
So we know the initial height of the balloon, which is the same as initiality of the rock.
10:46
So we know yb0 from the previous answer here.
10:57
And then we know the velocity of the balloon here.
11:04
That's our v0y.
11:08
So we can plug that in here.
11:18
So we know yb0 is just going to be y0 because the height, the initial height of the rock and the initial height of the balloon are the same.
11:27
So yb0 is equal to y0 and then y b0 is also the same as the initial y component of the rock.
11:42
Okay so that means y b or v b0 v b y is the same as v0 y.
12:00
Okay so then that gives us this equation here and now we can plug in y zero as you know that you plug in v b b b.
12:11
0y so we have this and we plug in the time t so it's going to give us the height the balloon has moved since the rock hit the ground so this say the balloon moves a height um this this height here um in the time it takes for the rock to the ground so this here is going to be your yb so now if we plug in the values i just mentioned into the yb equation here we're going to have a value of hundred meters.
13:09
Okay, so yb is 100 meters.
13:13
So now it's asking how high is the balloon when the rock hits the ground.
13:17
Okay.
13:18
So yb isn't going to necessarily give us the height of the balloon from the ground here.
13:23
It's just going to give us the change in height that it moved.
13:27
Okay, so in order to get the height here, let's see, i'll just call this yb prime.
13:41
This is going to give us the height of the balloon from the ground.
13:48
So in order to get this height, i have to take y0 minus yb to get yb prime.
13:54
And again, yb prime is the height of the balloon from the ground, which is what we're after.
14:15
So yb prime is equal to y0 minus yb.
14:35
So now i'm going to have, i just plug in the values for v0 and yb, or for y, 0, and yb.
14:46
So i'm going to get a height for the balloon of 123 meters.
14:52
All right, part c is asking at the ins and the rock hits the ground, how far is it from the basket? so now we want a distance x.
15:38
Okay.
15:40
So now we can use distance formula for x.
15:48
So i'm starting a new page now.
15:50
This is part c.
15:56
Okay, so we're now the distance x is equal to the initial velocity in the x direction...