A brush fire is burning on a rock ledge on one side of a ravine that is 20 m wide. A fire truck sits on the opposite side of the ravine at an elevation 6.5 m above the burning brush. The fire hose nozzle is aimed 35∘ above horizontal, and the firefighters control the water velocity by adjusting the water pressure. Because the water supply at a wilderness fire is limited, the firefighters want to use as little as possible.
At what speed should the stream of water leave the hose so that the water hits the fire on the first shot?
Damon's response is almost correct. Because the fire is 6.5m lower than the hose, you have to use -6.5 instead of 6.5 in the "h =" part. It causes the answer to end up being around 12 m/s instead.
To find the speed at which the stream of water should leave the hose, we can use the principle of projectile motion.
Step 1: Define the given variables:
- Width of the ravine: 20 m
- Elevation of the fire truck above the burning brush: 6.5 m
- Angle above the horizontal at which the nozzle is aimed: 35 degrees
Step 2: Split the motion into horizontal and vertical components.
- The horizontal component of the water's velocity does not change throughout the motion since there is no acceleration in that direction.
- The vertical component of the water's velocity changes due to gravity.
Step 3: Determine the time of flight for the water.
- The time it takes for the water to cross the ravine is the same as the time it takes for a projectile to fall a distance equal to the height difference between the fire truck and the burning brush.
- Using the equation for vertical motion: Δy = V₀y * t + (1/2) * g * t², where Δy is the vertical distance, V₀y is the initial vertical velocity, and g is the acceleration due to gravity (approximately 9.8 m/s²).
- Rearranging the equation, we get: t = √(2 * Δy / g)
Step 4: Calculate the initial velocity in the vertical direction.
- The vertical velocity can be determined using the equation V₀y = V₀ * sin(θ), where V₀ is the initial velocity and θ is the angle above horizontal.
- Rearranging the equation, V₀ = V₀y / sin(θ).
Step 5: Substitute the values into the equations and solve for the initial velocity, V₀.
- Δy = 6.5 m (height difference)
- θ = 35 degrees
- g = 9.8 m/s²
- Plugging in the values: t = √(2 * 6.5 / 9.8) and V₀ = V₀y / sin(35)
Step 6: Calculate the initial velocity in the horizontal direction.
- The horizontal velocity can be found using the equation V₀x = V₀ * cos(θ), where V₀x is the horizontal velocity.
- Since we are only interested in the speed, we can ignore the direction.
Step 7: Substitute the values into the equations and solve for the horizontal velocity, V₀x.
- θ = 35 degrees
- V₀x = V₀ * cos(35)
Step 8: Calculate the speed at which the stream of water should leave the hose.
- The speed is the magnitude of the velocity, so it can be calculated using the equation speed = √(V₀x² + V₀y²).
By following these steps, you will be able to find the speed at which the stream of water should leave the hose.
To determine the speed at which the stream of water should leave the hose so that it hits the fire on the first shot, we can break down the problem into two parts: the horizontal distance the water needs to travel to reach the fire and the vertical distance the water needs to rise to reach the fire.
Let's start by finding the horizontal distance the water needs to travel. We can use the concept of projectile motion to calculate this. The horizontal distance can be calculated using the formula:
d_horizontal = V_initial * t,
where V_initial is the initial horizontal velocity of the water and t is the time it takes for the water to reach the fire.
Since the horizontal speed of the water remains constant throughout its flight, we can denote V_horizontal as the horizontal velocity of the water.
Now, let's calculate the time it takes for the water to reach the fire. We can use the equation of motion in the vertical direction:
d_vertical = V_vertical_initial * t + (1/2) * a * t^2,
where V_vertical_initial is the initial vertical velocity of the water, a is the acceleration due to gravity (-9.8 m/s^2), and d_vertical is the vertical distance the water needs to rise.
The initial vertical velocity can be calculated using the formula:
V_vertical_initial = V_initial * sin(θ),
where θ is the angle at which the water is aimed above horizontal.
Since the water reaches its maximum height when it reaches the fire, the vertical displacement is equal to the elevation difference between the fire and the fire truck:
d_vertical = 6.5 m.
We also know that the horizontal distance is equal to the width of the ravine:
d_horizontal = 20 m.
Now, let's substitute the known values into the equations.
1. Vertical equation:
6.5 = (V_initial * sin(35∘)) * t + (1/2) * (-9.8 m/s^2) * t^2,
2. Horizontal equation:
20 = (V_initial * cos(35∘)) * t.
We have two equations and two unknowns (V_initial and t). We can solve these equations simultaneously to find the values.
Once we have the value of t, we can substitute it back into the horizontal equation to find V_initial.
Finally, we can find the speed at which the stream of water should leave the hose by calculating the magnitude of V_initial:
V = √(V_horizontal^2 + V_vertical_initial^2).
Solving these equations will provide the speed at which the stream of water should leave the hose.
s = that speed
u = s cos 35 = .819 s
Vi = s sin 35 = .574 s
20 = u t
so t = 20/u = 20/.819 s = 24.4/s
h = 6.5 = Vi t -4.9 t^2
6.5 = .574 s (24.4/s) - 4.9 (24.4/s)^2
6.5 = 14 - 2917 /s^2
2917 = 7.5 s^2
s = 19.7 m/s