A balloon carrying a basket is descending at a constant velocity of 20.0 m/s. A person in the basket throws a stone with an initial velocity of 15.0 m/s horizontally perpendicular to the path of the descending balloon, and 4.00s later this person sees the rock strike the ground.
A. How high was the balloon when the rock was thrown out?
B. How far horizontally does the rock travel before it hits the ground?
C. At the instant the rock hits the ground, how far is it from the basket?
A balloon carrying a basket is descending at a constant velocity of 20.0 m/s. A person in the basket throws a stone with an initial velocity of 15.0 m/s horizontally perpendicular to the path of the descending balloon, and 4.00s later this person sees the rock strike the ground.
A. How high was the balloon when the rock was thrown out?
From h = Vot + gt^2, where h = height, Vo = initial downward velocity, g = gravity = 9.8m/s and t = time in seconds, h = 20(4) + 9.8(4^2)/2 = 158.4m.
B. How far horizontally does the rock travel before it hits the ground?
Dh = 4(15) = 60met.
C. At the instant the rock hits the ground, how far is it from the basket?
The balloon descends 80m in 4 seconds being 78.4m from the ground at the rock's impact. Therefore, the distance between the balloon and the rock is sqrt(78.4^2 + 60^2) = 98.72 m.
Let's analyze the problem step by step:
Step 1: Determine the time it takes for the rock to hit the ground.
Given information:
- Initial velocity of the rock in the horizontal direction (Vx) = 15.0 m/s
- Time taken for the rock to hit the ground (t) = 4.00 s
Since the rock only experiences horizontal motion, we can use the formula:
Distance = Velocity * Time (d = v * t)
Using this formula, we can calculate the horizontal distance traveled by the rock:
Distance (horizontal) = Velocity (horizontal) * Time
Distance (horizontal) = 15.0 m/s * 4.00 s
Distance (horizontal) = 60.0 meters
So, the rock travels a horizontal distance of 60.0 meters before hitting the ground.
Step 2: Determine the height of the balloon when the rock was thrown out.
Given information:
- Velocity of the balloon (vertical) (Vy) = 20.0 m/s
- Time taken for the rock to hit the ground (t) = 4.00 s
Since the balloon is descending at a constant velocity, we can use the formula:
Distance = Velocity * Time (d = v * t)
Using this formula, we can calculate the vertical distance traveled by the balloon:
Distance (vertical) = Velocity (vertical) * Time
Distance (vertical) = 20.0 m/s * 4.00 s
Distance (vertical) = 80.0 meters
So, the balloon was 80.0 meters high when the rock was thrown out.
Step 3: Determine the horizontal distance between the rock and the basket when the rock hits the ground.
Given information:
- Initial velocity of the rock in the horizontal direction (Vx) = 15.0 m/s
- Time taken for the rock to hit the ground (t) = 4.00 s
- Velocity of the balloon (horizontal) (Vx) = 0 m/s (since it is not moving horizontally)
Since the rock and the basket have the same initial horizontal velocity and the balloon does not have any horizontal velocity, the horizontal distance between the rock and the basket remains constant.
Therefore, at the instant the rock hits the ground, the horizontal distance between the rock and the basket is the same as the horizontal distance traveled by the rock, which is 60.0 meters.
To summarize:
A. The balloon was 80.0 meters high when the rock was thrown out.
B. The rock travels a horizontal distance of 60.0 meters before hitting the ground.
C. At the instant the rock hits the ground, it is 60.0 meters away horizontally from the basket.
To answer these questions, we can break down the problem into its vertical and horizontal components.
A. How high was the balloon when the rock was thrown out?
To determine the height of the balloon when the rock was thrown out, we need to find the time it took for the rock to hit the ground since it was thrown. Given that the rock hits the ground after 4.00 seconds, and the balloon is descending at a constant velocity of 20.0 m/s, we can determine the altitude using the equation:
h = Δy = v * t
Where:
h = height (altitude)
v = velocity
t = time
Since the balloon is descending, the velocity is negative (-20.0 m/s). Plugging in the values, we get:
h = (-20.0 m/s) * (4.00 s)
h = -80.0 m
Therefore, the balloon was 80.0 meters above the ground when the rock was thrown out.
B. How far horizontally does the rock travel before it hits the ground?
Since the rock was thrown horizontally with an initial velocity of 15.0 m/s, and it takes 4.00 seconds to hit the ground, we can calculate the horizontal distance using the equation:
d = Δx = v * t
Where:
d = distance
v = velocity
t = time
In this case, the velocity is constant, but only in the horizontal direction. Plugging in the values, we get:
d = (15.0 m/s) * (4.00 s)
d = 60.0 m
Therefore, the rock travels 60.0 meters horizontally before it hits the ground.
C. At the instant the rock hits the ground, how far is it from the basket?
To determine the distance between the rock and the basket when it hits the ground, we need to consider the horizontal distance covered by the rock and the horizontal velocity of the balloon.
Given that the rock traveled 60.0 meters horizontally before hitting the ground, and the balloon has a constant horizontal velocity of 0 m/s (since it is descending vertically), we can conclude that the rock is 60.0 meters away from the basket when it hits the ground.