A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 25.2 m/s A 1.0-kg stone is thrown from the basket with an initial velocity of 11.7 m/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 6.70 s after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 25.2 m/s .How high was the balloon when the rock was thrown out?
How high is the balloon when the rock hits the ground?
At the instant the rock hits the ground, how far is it from the basket?
Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer at rest in the basket.
just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer at rest on the ground.
Oh boy, you've got quite the balloon adventure going on here! Let's dig into these questions with a touch of humor:
1. How high was the balloon when the rock was thrown out?
Well, it sounds like throwing a rock out of a balloon is always a good idea, right? To find the initial height, we need to use kinetic equations like a rockstar. Don't worry, we'll calculate it for you, no need to be a basket case!
2. How high is the balloon when the rock hits the ground?
Now, this is where things get interesting. Gravity is no laughing matter, it brings everything back down to Earth. So, as the rock plummets to the ground, the balloon's height will change, and not in a hot air way.
3. At the instant the rock hits the ground, how far is it from the basket?
When the rock finally makes its grand entrance on the ground, it will have traveled quite a distance from the basket (which I assume is along for the ride). Get ready for some mathematical excitement, we'll get you the exact distance!
4. Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer at rest in the basket.
Picture this: you're in the basket, sipping a refreshing beverage, when suddenly the rock takes off. It's like a rock-n-roll spectacle! We'll get you the horizontal and vertical velocity components before that rock rocks the ground!
5. Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer at rest on the ground.
Imagine being an observer on the ground, blissfully unaware of the rock's descent. But fear not, my friend! I'll help you calculate the horizontal and vertical velocity components just before the rock kisses the ground!
Now, all this talk of balloons, rocks, and velocities may seem a bit confusing, but fear not! I'll calculate all the answers for you. Remember, it's important to keep a smile on your face, even when dealing with physics.
To solve this problem, we can start by breaking it down into different parts and using the kinematic equations. Let's go step by step:
Step 1: Calculate the initial height of the balloon when the rock was thrown out.
We know that the downward velocity of the balloon is constant and equal to 25.2 m/s. Using this information, we can calculate the time it took for the stone to reach the ground by dividing the vertical distance (height of balloon) by the vertical velocity (downward velocity of the balloon).
Since the velocity is constant, we can use the equation: velocity = distance/time. Rearranging the equation, we have: time = distance/velocity.
Plugging in the values, we have: time = height of balloon / downward velocity of balloon = distance / 25.2 m/s.
The time it took for the stone to reach the ground is given as 6.70 seconds. So we can set up an equation: 6.70 s = distance / 25.2 m/s.
Solving for the distance, we have: distance = 6.70 s * 25.2 m/s = 168.84 m.
Therefore, the initial height of the balloon when the rock was thrown out is 168.84 meters.
Step 2: Calculate the height of the balloon when the rock hits the ground.
We already know the time it took for the stone to travel and hit the ground, which is 6.70 s. Using the constant downward velocity of the balloon (25.2 m/s), we can calculate the total distance the balloon traveled in that time.
The distance traveled by the balloon is given by: distance = velocity * time = 25.2 m/s * 6.70 s = 168.84 m.
Therefore, the height of the balloon when the rock hits the ground is also 168.84 meters.
Step 3: Calculate the distance between the basket and the spot where the rock hits the ground.
We can use the same equation as in step 2 to calculate the horizontal distance traveled by the balloon in 6.70 seconds, since the horizontal velocity of the balloon is unaffected by the stone's motion.
The horizontal distance is given by: distance = velocity * time = 25.2 m/s * 6.70 s = 168.84 m.
Therefore, the distance between the basket and the spot where the rock hits the ground is 168.84 meters.
Step 4: Calculate the horizontal and vertical components of velocity just before the rock hits the ground, as measured by an observer at rest in the basket.
Since the vertical velocity remains constant (downward at 25.2 m/s), the only change is in the horizontal velocity, due to the horizontal component of the stone's velocity.
The horizontal velocity component of the stone remains constant (11.7 m/s) throughout its flight. Therefore, the horizontal component of velocity just before the rock hits the ground, as measured by an observer at rest in the basket, is also 11.7 m/s.
The vertical velocity component remains constant at 25.2 m/s, as it was the initial velocity of the balloon.
Therefore, the horizontal velocity component just before the rock hits the ground is 11.7 m/s, and the vertical velocity component is 25.2 m/s.
Step 5: Calculate the horizontal and vertical components of velocity just before the rock hits the ground, as measured by an observer at rest on the ground.
To calculate the velocity components as measured by an observer at rest on the ground, we need to consider the horizontal and vertical components of the balloon's velocity.
The horizontal component of the balloon's velocity is not affected by the stone's motion, so it remains constant at 0 m/s.
The vertical component of the balloon's velocity remains constant at 25.2 m/s.
Therefore, the horizontal velocity component just before the rock hits the ground, as measured by an observer at rest on the ground, is 0 m/s, and the vertical velocity component is 25.2 m/s.
To find the height of the balloon when the rock was thrown out, we can use the concept of time and motion. We know that the person in the basket sees the rock hit the ground 6.70 s after being thrown. This means that the time it takes for the rock to hit the ground is the same as the time it took for the balloon to descend from its initial height to the current height.
Using the equation of motion for vertical motion, we can find the height of the balloon when the rock was thrown out. The equation is:
h = ut + (1/2)gt^2
where:
- h is the height of the balloon
- u is the initial vertical velocity (0 m/s as the balloon is descending with a constant downward velocity)
- g is the acceleration due to gravity (-9.8 m/s^2)
- t is the time taken for the descent (6.70 s)
Substituting the values, we get:
h = (0 m/s)(6.70 s) + (1/2)(-9.8 m/s^2)(6.70 s)^2
Therefore, the height of the balloon when the rock was thrown out is given by this equation.
To find the height of the balloon when the rock hits the ground, we can use the concept of vertical motion under constant acceleration. The equation for this situation is:
h = ht + (1/2)gt^2
where:
- h is the initial height of the balloon when the rock was thrown out
- ht is the height of the balloon when the rock hits the ground
- g is the acceleration due to gravity (-9.8 m/s^2)
- t is the time taken for the rock to hit the ground (6.70 s)
Substituting the values, we get:
0 = h + (1/2)(-9.8 m/s^2)(6.70 s)^2
Solving this equation will give us the height of the balloon when the rock hits the ground.
To find how far the rock is from the basket just before hitting the ground, we can use the concept of horizontal motion. The horizontal distance traveled by the rock can be calculated using:
d = vt
where:
- d is the horizontal distance traveled
- v is the horizontal velocity of the rock
- t is the time taken for the rock to hit the ground (6.70 s)
To find the horizontal velocity of the rock just before hitting the ground, we need to determine the change in horizontal velocity of the rock during its flight. Since no horizontal force acts on the rock, its horizontal velocity remains constant throughout its motion. Therefore, the horizontal velocity just before hitting the ground is the same as its initial horizontal velocity.
Finally, to find the horizontal and vertical velocity components of the rock just before hitting the ground, as measured by an observer at rest in the basket, we split the rock's velocity into horizontal and vertical components. The vertical component can be calculated using:
v_vertical = g * t
where:
- v_vertical is the vertical velocity of the rock just before hitting the ground
- g is the acceleration due to gravity (-9.8 m/s^2)
- t is the time taken for the rock to hit the ground (6.70 s)
The horizontal component of the velocity remains constant and can be found from the initial horizontal velocity of the rock.
To obtain the measurements as observed by an observer at rest on the ground, we need to consider the horizontal and vertical components of the rock's velocity separately. The horizontal component remains the same, while the vertical component will be equal to the final vertical velocity just before hitting the ground.