experiencing a constant horizontal 1.10 m/s wind, a hot air balloon ascends from the launch site at a constant vertical speed of 2.70 m/s. at a height of 202 m, the balloonist maintains constant altitude for 10.3 s before releasing a small sandbag. how far from the launch site will the sandbag land?
Are you to ignore the effect of wind on the sandbag? I assume so.
Time in air:
10.3+202/2.70 seconds
distance horizontal from launch :
1.1(timeinAir).
If it bag falls straight down, then that horizontal distance is it.
Now if the wind moves the sandbag, you have to add in time it takes to fall 202 meters (sqr(2*202/9.8) seconds, times the wind speed.
To determine the distance from the launch site where the sandbag will land, we need to calculate the horizontal distance it will travel during the 10.3 s it remains at a constant altitude.
Given:
Horizontal wind speed (Vh) = 1.10 m/s
Vertical speed (Vv) = 2.70 m/s
Height (h) = 202 m
Time at constant altitude (t) = 10.3 s
First, let's calculate the total time the balloon is in the air before the sandbag is released. Since the sandbag is released at a height of 202 m, we need to find the time it takes for the balloon to ascend to that height.
Use the formula:
h = Vv * t
202 m = 2.70 m/s * t
Solving for t:
t = 202 m / 2.70 m/s
t ≈ 74.81 s
Now, let's calculate the horizontal distance the balloon travels during this time.
Since the horizontal wind speed is constant, we can use the formula:
distance = Vh * time
distance = 1.10 m/s * 74.81 s
distance ≈ 82.29 m
So, the balloon will travel approximately 82.29 m horizontally in 74.81 s before releasing the sandbag.
Now, let's calculate how far the sandbag will travel horizontally during the 10.3 s it remains at a constant altitude.
distance = Vh * time
distance = 1.10 m/s * 10.3 s
distance ≈ 11.33 m
Therefore, the sandbag will land approximately 11.33 m horizontally away from the launch site.
To determine how far from the launch site the sandbag will land, we need to first analyze the horizontal and vertical motion of the sandbag.
Let's break down the problem into two parts: the time the sandbag is released and the time it takes for the sandbag to reach the ground.
1. Calculate the horizontal distance traveled by the sandbag while in the balloon:
Since the wind is blowing at a constant horizontal speed of 1.10 m/s and the balloon's vertical ascent doesn't affect the horizontal motion, we can consider the horizontal distance traveled by the sandbag to be the product of the constant wind speed and the time the sandbag is in the air.
The time before releasing the sandbag is given as 10.3 s. Therefore, the horizontal distance traveled by the sandbag while still in the balloon is:
Horizontal distance = Horizontal speed × Time
Horizontal distance = 1.10 m/s × 10.3 s = 11.33 m
2. Calculate the time it takes for the sandbag to fall:
When the sandbag is released, it will be affected by both the constant wind and gravitational acceleration.
The time taken for the sandbag to fall can be determined using the equation of motion:
Vertical distance = (Initial vertical velocity × Time) + (0.5 × Acceleration due to gravity × Time^2)
Since the sandbag is released from rest at a height of 202 m above the ground, the vertical distance is -202 m (negative because it is falling downward).
0 = (0 × Time) + (0.5 × 9.8 m/s^2 × Time^2)
0 = 4.9 m/s^2 × Time^2
Time^2 = 0
This implies that the time taken for the sandbag to fall is zero seconds since the equation reduces to zero.
3. Calculate the total horizontal distance traveled by the sandbag:
Since the time taken for the sandbag to fall is zero seconds, it means the sandbag will fall vertically straight down and will not be affected by the horizontal wind.
Therefore, the horizontal distance traveled by the sandbag is simply the horizontal distance it already traveled while in the balloon:
Horizontal distance = 11.33 m
Thus, the sandbag will land approximately 11.33 meters away from the launch site.