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

A 124 -kg balloon carrying a 22 -kg basket is descending with a constant downward velocity of 20.0 $\mathrm{m} / \mathrm{s} . \mathrm{A} 1.0$ -kg stone is thrown from the basket with an initial velocity of 15.0 $\mathrm{m} / \mathrm{s}$ perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. The person in the basket sees the stone hit the ground 6.00 s after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 $\mathrm{m} / \mathrm{s} .$ (a) How high was the balloon when the rock was thrown out? (b) How high is the balloon when the rock hits the ground? (c) At the instant the rock hits the ground, how far is it from the basket? (d) Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer (i) at rest in the basket and (ii) at rest on the ground.

Key Concepts

Uniform Motion

Uniform motion refers to motion at a constant velocity, meaning that the speed and direction of the object do not change over time. This concept is applied to the descending balloon in the problem, where the balloon’s vertical speed remains constant, making its displacement calculation straightforward.

Kinematics Equations

Kinematics equations provide mathematical relationships that describe the motion of objects under constant acceleration. These equations relate quantities such as displacement, initial velocity, time, and acceleration. They are crucial for solving problems involving both the uniformly accelerated vertical motion of a projectile and the constant horizontal motion.

Reference Frames

Reference frames are coordinate systems or perspectives from which motion is observed and measured. In dynamics, understanding the differences between observations made in different frames—such as a moving basket versus a stationary ground observer—is crucial for correctly applying the principles of relative velocity and kinematics.

Projectile Motion

Projectile motion refers to the two?dimensional motion of an object that is launched into the air and is subject only to acceleration due to gravity. This concept involves analyzing the motion by decomposing it into horizontal and vertical components, where the horizontal motion is at a constant velocity (in the absence of air resistance) and the vertical motion is uniformly accelerated motion under gravity.

Relative Velocity

Relative velocity deals with how the speed and direction of an object appear to change when observed from different reference frames. In problems involving projectiles thrown from moving platforms, the initial velocity relative to the platform has to be combined vectorially with the velocity of the platform to determine the velocity in a different, often stationary, frame of reference.

Transcript

00:01 This problem is really just checking to see if we understand some things about relative motion and vectors.

00:11 So there is a balloon like this with a basket like this.

00:23 And we'll just call this m big b.

00:29 I was going to write and we'll call this m little b and the whole thing is heading downwards like this.

00:46 We'll call this vb and it's the same for the balloon in the basket.

00:59 They're all moving together.

01:02 And then we have this stone.

01:07 I'm not drawn to scale here, but we're throwing this stone.

01:13 Out like that.

01:15 And this is the v s not in the x direction.

01:30 This is v b not in the y direction.

01:35 And this value here, they tell us that's 20 meters a second.

01:43 And for cleanliness, i'm going to drop some of the significant digits, but we know that it's out to the tenth's place and its accuracy.

01:51 And this is 15 meters a second.

01:59 But the important thing to note here is that the stone was in the basket.

02:06 So since the stone was in the basket, when we started, it had the motion of the basket.

02:12 It was in that frame of reference.

02:14 So the initial velocity of the stone in the y direction, was this same 20 meters a second.

02:33 And since these are the same vectors here, i'm going to go ahead and actually, i'm going to go ahead and make this a little bit shorter, just so it's kind of accurate.

02:47 This should be a little bit shorter than this vector because 15 is shorter than 20.

02:52 So hopefully that makes this a little bit more graphically accurate.

03:01 So this stone is going to move in some sort of a path kind of like this to the ground below.

03:13 So we want to find out a variety of things.

03:16 First thing is how high was the balloon when this rock was thrown out? so all we need to do is use an equation of motion.

03:28 Delta y equals v y not t plus one half g t squared and these are both going to be positive signs because i'm going to call down the positive direction the initial velocity of in the in the of the of the stone and the balloon are both downwards and acceleration is downwards which is why i pick down is the positive direction it's usually good to pick the direction of acceleration as the positive direction.

04:04 So they're both positive signs, and we can just plug in values here.

04:12 So this is going to be at 20, and we know that the time is six seconds.

04:24 There's our time.

04:26 So times six, and this is going to be meters.

04:31 Actually i'll go ahead and be specific with my units.

04:38 This is meters per seconds, that's seconds, plus one -half times 9 .8.

04:48 That's meters a second squared.

04:51 And this is six seconds squared.

04:58 So this is going to be 120 meters plus one -half times, six squared that's going to be 36 times one half that's 18 and times 9 .8 meters and if you put that in a calculator and do it all together we're going to get 296 .4 meters and if we were if this was just 10 here if we had rounded this to 10 it'd be 180 plus 120 that that'd be 300 meters.

05:48 So this is, you know, slightly more accurate if we use 9 .8, put around it to 10.

05:54 So this is the initial height, which is the initial height of the balloon and the stone.

06:01 That is this value here, delta y, here, from our starting point to the ground.

06:16 So then the next question here is how high is the balloon when the rock hits the ground? so this was our answer for a.

06:31 So then b, how high is when the rock hits the ground? well, this value, this part right here, this 120 meters, that was the initial velocity from the, that was the distance that the, that was the distance stone went due to its initial velocity, the 120 meters a second, that the balloon was descending.

07:07 So that's also the same distance that the balloon is going to descend, because the balloon continues to descend at the same exact speed, right? it keeps going down at the same constant speed of 20 meters a second it says right here.

07:25 So the balloon was 296 .4 meters above the ground.

07:32 So we know that delta y is going to equal y initial here we could say delta y b at if you wanted to do t equals six seconds if you want to be sort of really specific that's going to equal whatever why not was plus vb t or actually minus vbt because it's going to be a little a little bit less it's going to be less whatever this amount is so we were initially this 20096 .4 meters minus 20 meters a second times six seconds...

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