A person launches themselves into the air toward a landing which is 1.5m above ground level at a distance which is 4m from the release point of becoming airborne. What initial velocity should be used if a person swinging from a rope with an initial angle of 55 degrees, launched from 1m above ground Level? Explain the equations used and draw diagrams to solve this problem. 4m-
A person launches themselves into the air toward a landing which is 1.5m above ground level at a distance which is 4m from the release point of becoming airborne. What initial velocity should be used if a person swinging from a rope with an initial angle of 55 degrees, launched from 1m above ground Level? Explain the equations used and draw diagrams to solve this problem. 4m-
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Problem Description
A person launches themselves into the air toward a landing that is 1.5 meters above ground level at a distance of 4 meters from the release point of becoming airborne. What initial velocity should be used if a person is swinging from a rope with an initial angle of 55 degrees, launched from 1 meter above ground level? Explain the equations used and draw diagrams to solve this problem.
### Diagram Explanation
The diagram below illustrates the scenario:
- A person begins their motion at an angle \(\theta = 55^\circ\) above the horizontal.
- The initial launch point is 1 meter above the ground, depicted by a segment leading to the trajectory path.
- The trajectory forms an arc through the air ending at a landing platform which is 1.5 meters high.
- The horizontal distance from the launch to the landing is 4 meters.
### Solving the Problem
To determine the initial velocity (\(v_0\)), use the following physics equations:
#### Kinematic Equations
1. **Horizontal Motion**:
\[
x = v_0 \cdot \cos(\theta) \cdot t
\]
where \(x\) is the horizontal distance (4 m), \(v_0\) is the initial velocity, and \(t\) is the time of flight.
2. **Vertical Motion**:
\[
y = y_0 + v_0 \cdot \sin(\theta) \cdot t - \frac{1}{2} g t^2
\]
where \(y\) is the final vertical position (1.5 m), \(y_0\) is the initial vertical position (1 m), \(\theta\) is the launch angle, and \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)).
#### Steps
1. **Solve for Time (\(t\))**:
- From the horizontal motion equation:
\[
t = \frac{x}{v_0 \cdot \cos(\theta)}
\]
2. **Substitute into Vertical Motion Equation**:
- Replace \(t\) in the vertical motion equation and solve for \(v_0\).
By solving these equations simultaneously, you can find the required initial velocity \(v_0\) for the person to successfully](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2405e5cb-20d1-4277-a72a-e2ca429d5d5b%2Fe1baf181-487a-4fef-a7e6-3a61f2bd0383%2F9wm7nhp_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Description
A person launches themselves into the air toward a landing that is 1.5 meters above ground level at a distance of 4 meters from the release point of becoming airborne. What initial velocity should be used if a person is swinging from a rope with an initial angle of 55 degrees, launched from 1 meter above ground level? Explain the equations used and draw diagrams to solve this problem.
### Diagram Explanation
The diagram below illustrates the scenario:
- A person begins their motion at an angle \(\theta = 55^\circ\) above the horizontal.
- The initial launch point is 1 meter above the ground, depicted by a segment leading to the trajectory path.
- The trajectory forms an arc through the air ending at a landing platform which is 1.5 meters high.
- The horizontal distance from the launch to the landing is 4 meters.
### Solving the Problem
To determine the initial velocity (\(v_0\)), use the following physics equations:
#### Kinematic Equations
1. **Horizontal Motion**:
\[
x = v_0 \cdot \cos(\theta) \cdot t
\]
where \(x\) is the horizontal distance (4 m), \(v_0\) is the initial velocity, and \(t\) is the time of flight.
2. **Vertical Motion**:
\[
y = y_0 + v_0 \cdot \sin(\theta) \cdot t - \frac{1}{2} g t^2
\]
where \(y\) is the final vertical position (1.5 m), \(y_0\) is the initial vertical position (1 m), \(\theta\) is the launch angle, and \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)).
#### Steps
1. **Solve for Time (\(t\))**:
- From the horizontal motion equation:
\[
t = \frac{x}{v_0 \cdot \cos(\theta)}
\]
2. **Substitute into Vertical Motion Equation**:
- Replace \(t\) in the vertical motion equation and solve for \(v_0\).
By solving these equations simultaneously, you can find the required initial velocity \(v_0\) for the person to successfully
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