Homework Help is Here – Start Your Trial Now!
Science
Physics
A rifle that shoots bullets with a muzzle speed of 1461 m/s is aimed at a target 807.0 m away. If the center of the target is horizontally level with the rifle, at what height above the target must the rifle be pointed so that the bullet hits dead center? (a) 10.61 cm (b) 75.29 cm (c) 1.218 m (d) 1.495 m 15.

A rifle that shoots bullets with a muzzle speed of 1461 m/s is aimed at a target 807.0 m away. If the center of the target is horizontally level with the rifle, at what height above the target must the rifle be pointed so that the bullet hits dead center? (a) 10.61 cm (b) 75.29 cm (c) 1.218 m (d) 1.495 m 15.

College Physics
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)...
See similar textbooks
icon
Related questions
Question
Hello, I need help with question 15, the diagram is attached separately. Please and thank
**Projectile Motion Problem: Bullet Trajectory Calculation** **Problem Statement:** A rifle that shoots bullets with a muzzle speed of 1461 m/s is aimed at a target 807.0 m away. If the center of the target is horizontally level with the rifle, at what height above the target must the rifle be pointed so that the bullet hits dead center? **Options:** (a) 10.61 cm (b) 75.29 cm (c) 1.218 m (d) 1.495 m **Explanation:** In this problem, we are dealing with projectile motion, where a bullet is fired horizontally from a rifle. To solve this, we are required to calculate the vertical displacement of the bullet due to gravity while it travels horizontally to the target. This displacement will tell us the height above the target that the rifle should be aimed at so the bullet hits the center of the target. 1. **Calculate the time \( t \) it takes for the bullet to travel horizontally to the target:** Given: - Horizontal distance \( d = 807.0 \) m - Muzzle speed (horizontal speed) \( v_x = 1461 \) m/s Time \( t = \frac{d}{v_x} = \frac{807.0 \, \text{m}}{1461 \, \text{m/s}} = 0.5524 \) s 2. **Calculate the vertical displacement \( y \) due to gravity:** Using the formula for vertical displacement under constant acceleration (gravity), \( y = \frac{1}{2} g t^2 \), where: - \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)) - \( t = 0.5524 \) s \( y = \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (0.5524 \, \text{s})^2 = 1.495 \) m Therefore, the height above the target that the rifle must be pointed to hit the center is: (d) 1.495 m
Transcribed Image Text:**Projectile Motion Problem: Bullet Trajectory Calculation** **Problem Statement:** A rifle that shoots bullets with a muzzle speed of 1461 m/s is aimed at a target 807.0 m away. If the center of the target is horizontally level with the rifle, at what height above the target must the rifle be pointed so that the bullet hits dead center? **Options:** (a) 10.61 cm (b) 75.29 cm (c) 1.218 m (d) 1.495 m **Explanation:** In this problem, we are dealing with projectile motion, where a bullet is fired horizontally from a rifle. To solve this, we are required to calculate the vertical displacement of the bullet due to gravity while it travels horizontally to the target. This displacement will tell us the height above the target that the rifle should be aimed at so the bullet hits the center of the target. 1. **Calculate the time \( t \) it takes for the bullet to travel horizontally to the target:** Given: - Horizontal distance \( d = 807.0 \) m - Muzzle speed (horizontal speed) \( v_x = 1461 \) m/s Time \( t = \frac{d}{v_x} = \frac{807.0 \, \text{m}}{1461 \, \text{m/s}} = 0.5524 \) s 2. **Calculate the vertical displacement \( y \) due to gravity:** Using the formula for vertical displacement under constant acceleration (gravity), \( y = \frac{1}{2} g t^2 \), where: - \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)) - \( t = 0.5524 \) s \( y = \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (0.5524 \, \text{s})^2 = 1.495 \) m Therefore, the height above the target that the rifle must be pointed to hit the center is: (d) 1.495 m
The image above features a graph that appears to illustrate a comparison between linear and non-linear functions. Detailed descriptions of the features are as follows: 1. **Axes**: - The graph is set against a backdrop of a horizontal axis (x-axis) and a vertical axis (y-axis). - Both axes are marked with dotted lines beyond the graph for clarity. 2. **Curves and Lines**: - **Straight Line**: A dashed straight line runs diagonally from the bottom left to the upper right of the graph. This line represents a linear increase. - **Curved Line**: A smooth, concave curve starts from the same point as the straight line on the x-axis. The curved line rises initially but begins to dip, showing a non-linear relationship. 3. **Endpoints and Distances**: - Both lines extend until they intersect with a vertical line positioned on the right side of the graph. - A bracket marks the vertical distance between the endpoints of the two lines at this intersection point, with a question mark indicating that this distance is subject to evaluation or represents an unknown. 4. **Purpose**: - The graph visually demonstrates the difference in the y-values of a linear function and a non-linear function at a particular x-value. The bracketed distance at the vertical intersection suggests a comparison point, grappling with the difference or error between these two functions. In an educational context, such a graph can be used to teach concepts like linear versus non-linear growth, regression analysis, or the study of approximation errors in mathematical modeling. The graph prompts learners to engage with the idea of how different functions can diverge over an interval and highlights the importance of understanding these differences in various applications such as physics, economics, or engineering.
Transcribed Image Text:The image above features a graph that appears to illustrate a comparison between linear and non-linear functions. Detailed descriptions of the features are as follows: 1. **Axes**: - The graph is set against a backdrop of a horizontal axis (x-axis) and a vertical axis (y-axis). - Both axes are marked with dotted lines beyond the graph for clarity. 2. **Curves and Lines**: - **Straight Line**: A dashed straight line runs diagonally from the bottom left to the upper right of the graph. This line represents a linear increase. - **Curved Line**: A smooth, concave curve starts from the same point as the straight line on the x-axis. The curved line rises initially but begins to dip, showing a non-linear relationship. 3. **Endpoints and Distances**: - Both lines extend until they intersect with a vertical line positioned on the right side of the graph. - A bracket marks the vertical distance between the endpoints of the two lines at this intersection point, with a question mark indicating that this distance is subject to evaluation or represents an unknown. 4. **Purpose**: - The graph visually demonstrates the difference in the y-values of a linear function and a non-linear function at a particular x-value. The bracketed distance at the vertical intersection suggests a comparison point, grappling with the difference or error between these two functions. In an educational context, such a graph can be used to teach concepts like linear versus non-linear growth, regression analysis, or the study of approximation errors in mathematical modeling. The graph prompts learners to engage with the idea of how different functions can diverge over an interval and highlights the importance of understanding these differences in various applications such as physics, economics, or engineering.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 3 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Weakly interacting gases
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON
SEE MORE TEXTBOOKS