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Magazine Chapter 3: Problem 27
A fire hose ejects a stream of water at an angle of \(35.0^{\circ}\) above the horizontal. The water leaves the nozzle with a speed of \(25.0 \mathrm{m} / \mathrm{s}\). 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?
Short Answer
Expert verified
The fire hose should be placed approximately 59.86 meters away from the building.
Step by step solution
01
Decompose the Initial Velocity
To determine the range of the projectile, start by decomposing the initial velocity into horizontal and vertical components. The initial velocity is given as \(25.0 \ \mathrm{m/s}\) at an angle of \(35.0^{\circ}\). Calculate the horizontal component: \[ v_{0x} = v_0 \cos(35.0^{\circ}) = 25.0 \times \cos(35.0^{\circ}) \approx 20.5 \ \mathrm{m/s} \] Calculate the vertical component: \[ v_{0y} = v_0 \sin(35.0^{\circ}) = 25.0 \times \sin(35.0^{\circ}) \approx 14.3 \ \mathrm{m/s} \]
02
Calculate the Time to Reach Maximum Height
To find the time it takes for the water stream to reach its maximum height, use the vertical motion equation. At the maximum height, the vertical velocity component becomes zero:\[ v_{y} = v_{0y} - g t = 0 \]Solve for \(t\):\[ 0 = 14.3 - 9.81t \]\[ t = \frac{14.3}{9.81} \approx 1.46 \ \mathrm{s} \]
03
Calculate the Total Time of Flight
The total time of flight is twice the time to reach the maximum height, since the projectile takes equal time to ascend and descend:\[ t_{\text{total}} = 2 \times 1.46 \approx 2.92 \ \mathrm{s} \]
04
Calculate the Horizontal Distance
Now, calculate the horizontal distance traveled by the water stream during its flight using the horizontal component of the velocity:\[ x = v_{0x} \times t_{\text{total}} = 20.5 \times 2.92 \approx 59.86 \ \mathrm{m} \]
05
Conclusion
The fire hose should be placed approximately 59.86 meters away from the building to hit the highest point that the water can reach.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Velocity Decomposition
When dealing with projectile motion, the first step is to decompose the initial velocity into its components. The initial velocity is often provided as a single vector value and an angle. In our example, the water is ejected at 25.0 m/s at an angle of \(35.0^{\circ}\) above the horizontal.
- To find the horizontal component \(v_{0x}\), use the cosine of the angle: \(v_{0x} = v_0 \cos(35.0^{\circ})\).
- To find the vertical component \(v_{0y}\), use the sine of the angle: \(v_{0y} = v_0 \sin(35.0^{\circ})\).
Horizontal and Vertical Components
Understanding the horizontal and vertical components is crucial for analyzing projectile motion. The horizontal component, \(v_{0x}\), affects how far the projectile will travel horizontally, while the vertical component, \(v_{0y}\), dictates how high it can go.
- The horizontal motion is uniform since there is no acceleration (ignoring air resistance), and can be described as \(x = v_{0x} \cdot t\).
- The vertical motion is influenced by gravity, making it accelerate downwards. It can be described by the equation: \(v_{y} = v_{0y} - g \cdot t\).
Maximum Height Calculation
To calculate the maximum height in projectile motion, consider the vertical component. At the highest point, the vertical velocity equals zero, as gravity temporarily stops the ascent.
Using the equation \(v_{y} = v_{0y} - g \cdot t = 0\), you can solve for \(t\), the time to reach maximum height.
In this case, \(v_{0y} = 14.3 \ \mathrm{m/s}\), and \(g\) is the acceleration due to gravity (9.81 m/s虏). Plugging in the values gives \(t \approx 1.46 \ \mathrm{s}\).
Using the equation \(v_{y} = v_{0y} - g \cdot t = 0\), you can solve for \(t\), the time to reach maximum height.
In this case, \(v_{0y} = 14.3 \ \mathrm{m/s}\), and \(g\) is the acceleration due to gravity (9.81 m/s虏). Plugging in the values gives \(t \approx 1.46 \ \mathrm{s}\).
- This time allows calculating the maximum height using \(h = v_{0y} \cdot t - \frac{1}{2}gt^2\).
Time of Flight
The total time of flight for a projectile is crucial for determining how long it stays in the air. For symmetric projectiles like this stream of water, where it lands at the same height it was launched, the total flight time is twice the time to reach the maximum height.
Since the time to reach the top was calculated as 1.46 seconds, the total time of flight \(t_{\text{total}}\) is \(2 \times 1.46 \ \approx 2.92 \ \mathrm{s}\).
Since the time to reach the top was calculated as 1.46 seconds, the total time of flight \(t_{\text{total}}\) is \(2 \times 1.46 \ \approx 2.92 \ \mathrm{s}\).
- This duration considers both ascending and descending phases of the projectile's journey.
Horizontal Distance Calculation
The horizontal distance a projectile travels is determined by the horizontal component of its initial velocity and the total time of flight. This is calculated using the equation \(x = v_{0x} \cdot t_{\text{total}}\).
Given \(v_{0x} = 20.5 \ \mathrm{m/s}\) and \(t_{\text{total}} \approx 2.92 \ \mathrm{s}\), the distance \(x\) can be found by multiplying these values, resulting in approximately 59.86 meters.
Given \(v_{0x} = 20.5 \ \mathrm{m/s}\) and \(t_{\text{total}} \approx 2.92 \ \mathrm{s}\), the distance \(x\) can be found by multiplying these values, resulting in approximately 59.86 meters.
- The fire hose should be positioned about 59.86 meters away from a building to ensure the water reaches the highest possible point.
