/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Firemen are shooting a stream of... [FREE SOLUTION] | 91Ó°ÊÓ

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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 s to reach a building 45.0 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 m/s, Acceleration: 9.8 m/s², (c) Height: 30.9 m, Speed: 17.8 m/s.

Step by step solution

01

Break down the projectile motion components

In projectile motion, the horizontal and vertical components can be analyzed independently. The initial speed \( v_0 = 25.0 \text{ m/s} \) must be split into horizontal and vertical components using the angle \( \alpha \): \[ v_{0x} = v_0 \cos(\alpha) \] and \[ v_{0y} = v_0 \sin(\alpha) \].
02

Set up horizontal motion equation

The time of flight \( t = 3.00 \text{ s} \) and horizontal distance \( x = 45.0 \text{ m} \) relate through the equation: \[ x = v_{0x} \cdot t \]. Thus, \[ 45.0 = 25.0 \cos(\alpha) \cdot 3.00 \]. From this equation, we can solve for \( \cos(\alpha) \): \[ \cos(\alpha) = \frac{45.0}{75.0} = 0.60 \].
03

Use trigonometry to find the angle \( \alpha \)

Given \( \cos(\alpha) = 0.60 \), use the inverse cosine function to find the angle \( \alpha \): \[ \alpha = \cos^{-1}(0.60) \approx 53.13^\circ \].
04

Vertical motion at the peak

At the highest point of the trajectory, the vertical component of velocity \( v_{0y} \) becomes zero. Thus, the speed of the water at the peak is equal to the horizontal component, \( v_{0x} = 25.0 \cos(\alpha) = 15 \text{ m/s} \). The only acceleration acting is gravity, \( 9.8 \text{ m/s}^2 \), in the downward direction.
05

Calculate vertical distance at impact with building

Since the time to reach the building is \( 3.00 \text{ s} \), use the vertical motion equation: \[ y = v_{0y} \cdot t - \frac{1}{2}gt^2 \]. From Step 1, \( v_{0y} = 25.0 \sin(\alpha) \approx 20 \text{ m/s} \). Thus, \[ y = 20 \cdot 3.00 - \frac{1}{2} \cdot 9.8 \cdot 3.00^2 \approx 30.9 \text{ m} \].
06

Find speed before hitting the building

The speed of the water just before impact combines both horizontal and vertical components. At impact, \( v_y \) is derived from \( v_{0y} - gt \): \[ v_y = 20 - 9.8 \cdot 3 = -9.4 \text{ m/s} \] and \( v_x = 15 \text{ m/s} \). The total speed \( v \) is: \[ v = \sqrt{v_x^2 + v_y^2} \approx \sqrt{15^2 + (-9.4)^2} \approx 17.8 \text{ m/s} \].

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause the motion. In this exercise, the water stream from the hose exhibits projectile motion, a type of motion experienced by an object thrown or projected into the air, subject only to the acceleration of gravity. The key to solving this type of problem is to analyze both the horizontal and vertical components of the projectile separately, which lends itself to an easier understanding and solution.

Kinematics involves using several basic equations that describe displacement, velocity, and acceleration as a function of time. These equations are simplified because, in projectile motion, the only acceleration acting is gravity, which impacts the vertical component, while the horizontal component moves with constant velocity. This separation of motion into two independent components is crucial for breaking down complex trajectories into solvable equations.
Angle of Elevation
The angle of elevation is the angle formed between the horizontal line and the line of sight to an object above that horizontal line. In this exercise, the angle of elevation \( \alpha \) is a key factor in determining how far and high the water stream will travel. Adjusting this angle changes the balance of horizontal and vertical speeds, which in turn affects the projectile's range and height.

To find the angle of elevation, use trigonometry with the known values of horizontal distance and initial speed. Using the cosine function, the horizontal component of the velocity is related to the total velocity and angle through \( v_{0x} = v_0 \cos(\alpha) \). Solving for \( \alpha \) provides the precise angle needed to reach a particular point with a fixed time of flight, as demonstrated in the step-by-step solution.
Horizontal and Vertical Components
In projectile motion, the initial velocity can be divided into horizontal and vertical components using trigonometric functions. These components allow us to independently examine the projectile's path in both dimensions. For this specific problem, the initial velocity is given as 25.0 m/s.

By applying the angle of elevation, we determine:
  • Horizontal component: \( v_{0x} = v_0 \cos(\alpha) \)
  • Vertical component: \( v_{0y} = v_0 \sin(\alpha) \)
These components help solve for additional parameters like time of flight, maximum height, and the point of impact. With the equation \( x = v_{0x} \cdot t \) for horizontal motion, and \( y = v_{0y} \cdot t - \frac{1}{2}gt^2 \) for vertical motion, it becomes possible to calculate crucial aspects of projectile motion with clear, focused calculations.
Trajectory Analysis
Trajectory analysis involves understanding the complete path taken by the projectile, from the point of launch to the point of impact. Using the principal components explored earlier, one can piece together the projectile's path.

Key points to analyze include:
  • The peak of the trajectory, where vertical velocity is zero.
  • Point of impact, determining horizontal distance traveled and vertical displacement at the end of motion.
  • Speed at different points (most notably, at launch, at the peak, and just before impact).
Through trajectory analysis, we can answer critical questions like how high the water stream travels and its speed when hitting the building. As shown in the detailed calculations, each part of the path has its own characteristics that can be calculated and understood using kinematics principles.

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