A fire hose ejects a stream of water at an angle of 35.0º above the horizontal. The water leaves the nozzle with a speed of 25.0 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?
25ft
30m
29.7 m
... solve for X by finding the time it takes for v0y to hit 0
1.45s times 20.5 m/s (vx) = 29.7 meters
To determine how far from the building the fire hose should be located to hit the highest possible fire, we can start by breaking down the problem into horizontal and vertical components.
Given:
- Angle of elevation, θ = 35.0º
- Initial speed of water, v = 25.0 m/s
1. First, let's find the initial vertical velocity component (vy) and the initial horizontal velocity component (vx). We can use trigonometry to solve for these values.
vy = v * sin(θ)
vx = v * cos(θ)
Substituting the given values:
vy = 25.0 m/s * sin(35.0º)
vx = 25.0 m/s * cos(35.0º)
Calculating vy and vx:
vy ≈ 14.3 m/s
vx ≈ 20.4 m/s
2. Next, let's determine the time it takes for the water to reach its peak height. At the peak, the vertical velocity component will be zero (vy = 0). We can use the kinematic equation:
vy = vy0 + gt
where vy0 is the initial vertical velocity component and g is the acceleration due to gravity (-9.8 m/s²).
Rearranging the equation:
t = -vy0 / g
Substituting the values:
t = -14.3 m/s / (-9.8 m/s²)
Calculating t:
t ≈ 1.46 s
3. Now, we can find the horizontal distance traveled by the water during this time.
Using the equation:
d = vx * t
Substituting the values:
d = 20.4 m/s * 1.46 s
Calculating d:
d ≈ 29.8 m
Therefore, the fire hose should be located approximately 29.8 meters away from the building to hit the highest possible fire.