A fire hose ejects a stream of water at an angle of 31.5 ° above the horizontal. The water leaves the nozzle with a speed of 24.3 m/s. Assuming that the water behaves like a projectile, how far from a building should the fire hose be located to hit the highest possible fire?
If the hose elevation angle remains 31.5 degrees, the maximum height obtained for the water stream is
H = (Vo sin31.5)^2/(2g)
= 8.22 m
The stream reaches that height when the distance from the building is
H cos31.5 = 7.01 m
To determine the distance from the building where the water will hit the highest possible fire, we need to find the range of the water stream.
Given:
- Angle of elevation (θ) = 31.5 °
- Initial velocity (v) = 24.3 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
The range (R) of a projectile is given by the formula:
R = (v² * sin(2θ)) / g
Let's substitute the given values into the formula:
R = (24.3² * sin(2 * 31.5)) / 9.8
R ≈ (591.249 * sin(63)) / 9.8
R ≈ (591.249 * 0.891) / 9.8
R ≈ 56.749 / 9.8
R ≈ 5.8 meters
Therefore, the fire hose should be located about 5.8 meters away from the building to hit the highest possible fire.
To determine how far from the building the fire hose should be located to hit the highest possible fire, we will analyze the water stream as a projectile motion problem.
First, let's break down the given information:
- Initial velocity (launch speed): v₀ = 24.3 m/s
- Launch angle: θ = 31.5° above the horizontal
- Acceleration due to gravity: g = 9.8 m/s² (assuming no air resistance)
We need to find the horizontal distance (range) traveled by the water stream before hitting the highest possible fire.
To solve this, we can use the following projectile motion equations:
1. The horizontal component of velocity (v₀x):
v₀x = v₀ * cos(θ)
2. The time it takes for the water stream to reach its highest point (t):
t = (v₀ * sin(θ)) / g
3. The range of the water stream (R):
R = v₀x * t
Let's calculate the values:
1. v₀x = v₀ * cos(θ)
v₀x = 24.3 m/s * cos(31.5°)
v₀x ≈ 20.924 m/s
2. t = (v₀ * sin(θ)) / g
t = (24.3 m/s * sin(31.5°)) / 9.8 m/s²
t ≈ 0.412 s
3. R = v₀x * t
R = 20.924 m/s * 0.412 s
R ≈ 8.618 m
Therefore, the fire hose should be located approximately 8.618 meters away from the building to hit the highest possible fire.