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Firemen are shooting a stream of water at a burning building using a high- pressure hose that shoots out the water with a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) as it leaves the end of the hose. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation \(\alpha\) of the hose until the water takes 3.00 \(\mathrm{s}\) to reach a building 45.0 \(\mathrm{m}\) away. You can ignore air resistance; assume that the end of the hose is at ground level. (a) Find the angle of elevation \(\alpha .\) (b) Find the speed and acceleration of the water at the highest point in its trajectory. (c) How high above the ground does the water strike the building, and how fast is it moving just before it hits the building?

Short Answer

Expert verified
(a) \(\alpha \approx 53.13^\circ\). (b) Speed = 15.0 m/s; Acceleration = 9.81 m/s² down. (c) Height = 5.86 m; Speed = 17.57 m/s.

Step by step solution

01

Understand the Problem

We are given: - Initial speed of water, \( v_0 = 25.0 \, \mathrm{m/s} \)- Time of flight to reach the building, \( t = 3.00 \, \mathrm{s} \)- Horizontal distance to the building, \( d_x = 45.0 \, \mathrm{m} \).We need to find:1. The angle of elevation \( \alpha \).2. The speed and acceleration of the water at the highest point.3. The height above the ground where the water strikes the building and its speed at this point. We can ignore air resistance throughout.
02

Break down the motion into components

Split the initial velocity into horizontal and vertical components:- Horizontal component: \( v_{0x} = v_0 \cos \alpha \)- Vertical component: \( v_{0y} = v_0 \sin \alpha \)The motion is a projectile, so these components will help in solving the problem.
03

Find the angle of elevation \( \alpha \)

For horizontal motion, use the formula:\[ d_x = v_{0x} \times t = v_0 \cos \alpha \times t \]Rearrange to solve for \( \alpha \):\[ \cos \alpha = \frac{d_x}{v_0 t} = \frac{45.0}{25.0 \times 3.0} = 0.6 \]Finally, find \( \alpha \):\[ \alpha = \cos^{-1}(0.6) \approx 53.13^\circ \]
04

Find the speed and acceleration at the highest point

At the highest point in its trajectory, the vertical component of the velocity is zero. Thus:- Speed at the highest point is purely horizontal: \( v_{highest} = v_{0x} = v_0 \cos \alpha = 25.0 \times 0.6 = 15.0 \, \mathrm{m/s} \)- Acceleration is only due to gravity: \( a = g = 9.81 \, \mathrm{m/s^2} \) directed downwards.
05

Determine the height at which water strikes the building

For vertical motion:Using the kinematic equation:\[ y = v_{0y} t - \frac{1}{2} g t^2 \]Find \( y \):\[ v_{0y} = v_0 \sin \alpha = 25.0 \times \sin(53.13^\circ) \approx 20.0 \, \mathrm{m/s} \]\[ y = 20.0 \times 3.0 - \frac{1}{2} \times 9.81 \times 3.0^2 = 30.0 - 44.145 \approx 5.855 \, \mathrm{m} \]
06

Calculate the speed just before hitting the building

The velocity before hitting the building:- Horizontal: \( v_{x} = v_{0x} = 15.0 \, \mathrm{m/s} \)- Vertical from kinematic equations, \( v_y = v_{0y} - g \times t = 20.0 - 9.81 \times 3.0 = -9.43 \, \mathrm{m/s} \) \( (\text{since the water is falling downwards)} \)Use Pythagoras to find magnitude of velocity:\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{15.0^2 + (-9.43)^2} \approx 17.57 \, \mathrm{m/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angle of Elevation
The angle of elevation is crucial in projectile motion as it determines the initial direction of the projectile. In this scenario, it refers to the angle at which the water is being shot from the hose. This is a critical variable because it influences how far and how high the water can travel. To find the angle of elevation, you can use the relationship between horizontal distance, initial velocity, and time.
For the water to reach the burning building 45 meters away, the angle of elevation must be set so that the horizontal component of velocity can cover this distance in the given time of 3 seconds. By applying the formula \ \( d_x = v_0 \cos \alpha \times t \ \), and knowing that \ \( d_x \) is 45 m, \ \( v_0 \) is 25 m/s, and \ \( t \) is 3 s, we can calculate the angle \( \alpha \) to be approximately 53.13 degrees. This ensures the water stream reaches the target effectively.
Kinematic Equations
Kinematic equations are a set of equations that describe the motion of objects without considering the forces that cause this motion. They are essential in solving projectile motion problems. These equations relate the five key kinematic variables: displacement, initial velocity, final velocity, acceleration, and time.
In our exercise, they help determine the trajectory and final outcomes of the water stream after it is projected. For the vertical rise and fall of the water, the kinematic equation \ \( y = v_{0y} \times t - \frac{1}{2} g t^2 \ \) helps calculate how high the stream goes above the ground. Similarly, we use \ \( v = v_{0y} - g \times t \ \) to calculate the vertical component of velocity before the water hits the building. Understanding how these equations apply is essential, as they enable us to predict aspects like the highest point of travel and the impact speed against the building.
Horizontal and Vertical Components
In projectile motion, it's beneficial to break the initial velocity into horizontal and vertical components. This simplification makes calculations manageable and helps understand the trajectory better.
The horizontal component, calculated as \ \( v_{0x} = v_0 \cos \alpha \ \), influences how far the projectile travels horizontally. For our water stream, this determines whether it reaches the building. The vertical component, \ \( v_{0y} = v_0 \sin \alpha \ \), is responsible for the ascent and descent of the projectile. It dictates how high the water will rise before gravity pulls it back down. At the highest point, the vertical velocity is zero, meaning the water momentarily stops moving upwards before descending.
Splitting these components helps solve for different variables in projectile motion, such as how to find the height the water hits the building, or its speed at any given point.
Speed and Acceleration
Speed refers to how fast something is moving, while acceleration describes the rate at which this speed changes. During projectile motion, speed and acceleration vary at different points along the path.
At the highest point of a projectile's path, the vertical component of the velocity becomes zero, but the horizontal component remains unchanged because there's no horizontal acceleration (neglecting air resistance). Therefore, the speed at this point is purely horizontal. For our water stream, this speed is 15 m/s purely in the horizontal direction.
Acceleration is constant throughout at 9.81 m/s² downwards (due to gravity). This affects the vertical velocity of the water, causing it to decrease as it rises and then increases in magnitude (but in the downward direction) as it falls back down. Understanding these changes helps us determine critical outcomes, like how fast and at what angle the water will hit the building.

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Most popular questions from this chapter

When it is 145 \(\mathrm{m}\) above the ground, a rocket traveling vertìcally upward at a constant 8.50 \(\mathrm{m} / \mathrm{s}\) relative to the ground launches a secondary rocket at a speed of 12.0 \(\mathrm{m} / \mathrm{s}\) at an angle of \(53.0^{\circ}\) above the horizontal, both quantities being measured by an astronaut sitting in the rocket. Air resistance is too small to worry about. (a) Just as the secondary rocket is lunched, what are the horizontal and vertical components of its velocity relative to (i) the astronaut sitting in the rocket and (ii) Mission Control on the ground? (b) Find the initial speed and launch angle of the secondary rocket as measured by Mission Control. (c) What maximum height above the ground does the secondary rocket reach?

(a) Prove that a projectile launched at angle \(\alpha_{0}\) has the same horizontal range as one launched with the same speed at angle \(\left(90^{\circ}-\alpha_{0}\right) \cdot(\text { b) A frog jumps at a speed of } 2.2 \mathrm{m} / \mathrm{s} \text { and lands } 25 \mathrm{cm}\) from its starting point. At which angles above the horizontal could it have jumped?

Tossing Your Lunch. Henrietta is going off to her physics class, jogging down the sidewalk at 3.05 \(\mathrm{m} / \mathrm{s}\) . Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is 43.9 \(\mathrm{m}\) above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally 9.00 \(\mathrm{s}\) after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance. (a) With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground? (b) Where is Henrietta when she catches the bagels?

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