A fire hose ejects a stream of water at an angle of 29.0° above the horizontal. The water leaves the nozzle with a speed of 31.0 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?
To determine the distance from the building at which the fire hose should be located to hit the highest possible fire, we need to find the horizontal distance covered by the water stream.
First, let's find the time of flight for the water stream. We can use the vertical motion equation:
y = y₀ + v₀y * t - 1/2 * g * t²
Since the water stream reaches its highest point, the final vertical displacement (y) will be zero. The initial vertical displacement (y₀) is also zero because the water stream starts from the ground. The initial vertical velocity (v₀y) is given by:
v₀y = v₀ * sin(θ)
where v₀ is the initial speed of the water stream and θ is the launch angle.
Thus, the equation becomes:
0 = 0 + (31.0 m/s * sin(29.0°)) * t - 1/2 * 9.8 m/s² * t²
Simplifying the equation, we get:
4.9 * t² = 15.0 * sin(29.0°) * t
Now, let's solve the equation for t:
4.9 * t = 15.0 * sin(29.0°)
t = (15.0 * sin(29.0°)) / 4.9
t ≈ 1.4687 seconds
Next, we can find the horizontal distance covered by the water stream using the horizontal motion equation:
x = v₀x * t
where v₀x is the initial horizontal velocity. Since there is no horizontal acceleration, v₀x remains constant throughout the motion.
To find v₀x, we can use the launch angle θ and the initial speed v₀:
v₀x = v₀ * cos(θ)
v₀x = 31.0 m/s * cos(29.0°)
v₀x ≈ 27.5008 m/s
Now, we can calculate the horizontal distance x:
x = 27.5008 m/s * 1.4687 seconds
x ≈ 40.414 meters
Therefore, the fire hose should be located approximately 40.414 meters away from the building to hit the highest possible fire.
To find the distance from the building where the water hits the highest possible fire, we need to analyze the projectile motion of the water stream ejected from the fire hose.
The given information states that the water stream leaves the nozzle with a speed of 31.0 m/s at an angle of 29.0° above the horizontal. We can break down the initial velocity into its horizontal and vertical components.
The horizontal component of the initial velocity (Vx) can be calculated using the equation:
Vx = V * cos(theta)
where V is the initial speed of the water stream and theta is the angle above the horizontal. Plugging in the given values:
Vx = 31.0 m/s * cos(29.0°)
Vx ≈ 27.53 m/s
The vertical component of the initial velocity (Vy) can be calculated using the equation:
Vy = V * sin(theta)
where V is the initial speed of the water stream and theta is the angle above the horizontal. Plugging in the given values:
Vy = 31.0 m/s * sin(29.0°)
Vy ≈ 15.19 m/s
Now, we need to find the time it takes for the water stream to reach the highest point of its trajectory. At the highest point, the vertical velocity becomes zero.
Using the equation for vertical velocity at any given time:
Vy = Vy0 + a * t
where Vy is the vertical velocity, Vy0 is the initial vertical velocity, a is the acceleration due to gravity (-9.8 m/s²), and t is time.
Plugging in the values:
0 = 15.19 m/s + (-9.8 m/s²) * t
Solving for t:
t = 15.19 m/s / 9.8 m/s²
t ≈ 1.55 s
Now we can find the distance the water stream travels horizontally in this time period. The horizontal distance (d) can be calculated using the equation:
d = Vx * t
Plugging in the values:
d = 27.53 m/s * 1.55 s
d ≈ 42.64 m
Therefore, the fire hose should be located approximately 42.64 meters away from the building to hit the highest possible fire.