A brush fire is burning on a rock ledge on one side of a ravine that is 35 m wide. A fire truck sits on the opposite side of the ravine at an elevation 4.5 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.
Part A
At what speed should the stream of water leave the hose so that the water hits the fire on the first shot?
Horizontal distance: 35=Vcos35*t
vertical ditance: hf=ho+Vsin35*t-1/2 g t^2
solve for t in the first equation: t= 35/VCos35=42.7/V put it in for t in the second equation:
0=4.5+Vsin35(42.7/V)-4.9(42.7/V)^2
multiply thru by V^2, and you have a quadratic. Use the quadratic formula to slove for V
To solve this problem, we need to find the speed at which the water should leave the hose so that it reaches the fire on the first shot. We can use the principles of projectile motion to solve this.
Given information:
- Ravine width (d): 35 m
- Elevation of fire truck (h): 4.5 m
- Angle of nozzle above horizontal (θ): 35°
- Vertical displacement (Δy): h (elevation of fire truck)
- Horizontal displacement (Δx): d (ravine width)
We can break down the initial velocity of the water into its vertical and horizontal components:
Vertical component: V₀y = V₀ * sin(θ)
Horizontal component: V₀x = V₀ * cos(θ)
Let's start by calculating the time it takes for the water to reach the other side of the ravine (horizontal displacement):
Δx = V₀x * t
35m = V₀ * cos(35°) * t [Equation 1]
Next, let's calculate the time it takes for the water to reach the same elevation as the fire (vertical displacement):
Δy = V₀y * t + (1/2) * g * t^2
4.5m = V₀ * sin(35°) * t - (1/2) * g * t^2 [Equation 2]
We need to solve these two equations simultaneously to find the initial velocity (V₀):
From Equation 1:
V₀ = 35m / (cos(35°) * t)
Substitute V₀ into Equation 2:
4.5m = (35m / (cos(35°) * t)) * sin(35°) * t - (1/2) * g * t^2
Simplifying:
4.5m = (35m * sin(35°)) / cos(35°) - (1/2) * g * t^2
Let's solve this equation for t:
t^2 = (70m * sin(35°)) / (g * cos(35°)) - (9m / (g * cos(35°)))
t ≈ √((70m * sin(35°)) / (g * cos(35°)) - (9m / (g * cos(35°))))
Now, substitute the value of t into Equation 1 to find V₀:
V₀ = 35m / (cos(35°) * (√((70m * sin(35°)) / (g * cos(35°)) - (9m / (g * cos(35°))))))
Finally, we can use the value of g (acceleration due to gravity) which is approximately 9.8 m/s² to calculate the numerical value.
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 given information and use principles of projectile motion.
Let's start by visualizing the situation. We have a right triangle formed by the ravine, with the horizontal distance (across the ravine) between the fire truck and the fire being 35 m, and the vertical distance (elevation) between the fire truck and the fire being 4.5 m. The water stream leaves the hose at an angle of 35 degrees above the horizontal.
To find the speed at which the water stream should leave the hose, we need to analyze the horizontal and vertical components of the water's velocity separately.
1. Horizontal Component:
The horizontal component of velocity remains constant throughout the motion because there is no horizontal acceleration (assuming air resistance is negligible). The only horizontal force acting on the water stream is due to Earth's gravity, which does not affect the horizontal component of velocity. Therefore, the horizontal component of velocity will be the same as the initial horizontal speed.
2. Vertical Component:
The vertical component of velocity will change due to the acceleration caused by gravity. The vertical acceleration is equal to the acceleration due to gravity, which is approximately 9.8 m/s^2 downward.
Now, let's find the initial speed of the water stream:
1. Horizontal Component:
The horizontal velocity (Vx) can be found using the equation:
Vx = V * cos(theta)
where V is the initial speed of the water stream and theta is the angle of elevation (35 degrees in this case).
2. Vertical Component:
The vertical velocity (Vy) can be found using the equation:
Vy = V * sin(theta)
The time taken by the water stream to reach the burning brush is the same for both horizontal and vertical components because they are launched simultaneously. Therefore, we only need to consider the horizontal distance (35 m) and its corresponding horizontal velocity. We can calculate the time (t) using the equation:
distance = velocity * time
35 m = Vx * t
t = 35 m / Vx
Since the vertical distance is not provided, we will solve for Vx and substitute it into the equation to find the water's horizontal speed:
t = 35 m / (V * cos(theta))
The goal is to hit the fire on the first shot, which means the water stream should reach the fire at the same time it takes for the water to cross the ravine horizontally.
Now, let's solve for the initial speed of the water stream (V):
V = distance / (time * cos(theta))
V = 35 m / (35 m / (V * cos(theta)) * cos(theta))
V = 35 m / (35 m / cos(theta))
V = 1 / cos(theta)
V = 1 / cos(35 degrees)
Using a calculator, we find V ≈ 1.217
Therefore, the stream of water should leave the hose at a speed of approximately 1.217 m/s to hit the fire on the first shot.