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Which diagram represents an accurate sketch of the rock's trajectory? View Available Hint(s) о k U₁ 15 15 18 18 d Uf to Uf U₁

Which diagram represents an accurate sketch of the rock's trajectory? View Available Hint(s) о k U₁ 15 15 18 18 d Uf to Uf U₁

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)...
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**Learning Goal:** A rock thrown with speed 8.50 m/s and launch angle 30.0° (above the horizontal) travels a horizontal distance of \( d = 19.0 \, \text{m} \) before hitting the ground. From what height was the rock thrown? Use the value \( g = 9.800 \, \text{m/s}^2 \) for the free-fall acceleration. --- **PROBLEM-SOLVING STRATEGY 4.1: Projectile motion problems** **MODEL:** Is it reasonable to ignore air resistance? If so, use the projectile motion model. **VISUALIZE:** Establish a coordinate system with the x-axis horizontal and the y-axis vertical. Define symbols and identify what the problem is trying to find. For a launch at angle \( \theta \), the initial velocity components are \( v_{ix} = v \cos \theta \) and \( v_{iy} = v \sin \theta \). **SOLVE:** The acceleration is known: \( a_x = 0 \) and \( a_y = -g \). Thus, the problem becomes one of two-dimensional kinematics. The kinematic equations are \[ \begin{array}{lc} \text{Horizontal} & \text{Vertical} \\ x_f = x_i + v_{ix} \Delta t & y_f = y_i + v_{iy} \Delta t - \frac{1}{2} g (\Delta t)^2 \, , \\ v_{tx} = v_{ix} = \text{constant} & v_{ty} = v_{iy} - g \Delta t \, \end{array} \] \( \Delta t \) is the same for the horizontal and vertical components of the motion. Find \( \Delta t \) from one component, and then use that value for the other component. **REVIEW:** Check that your result has the correct units and significant figures, is reasonable, and answers the question. --- **Model** Start by making simplifying assumptions: Model the rock as a particle in free fall. You can ignore air resistance because the rock is a relatively heavy object moving relatively slowly.
Transcribed Image Text:**Learning Goal:** A rock thrown with speed 8.50 m/s and launch angle 30.0° (above the horizontal) travels a horizontal distance of \( d = 19.0 \, \text{m} \) before hitting the ground. From what height was the rock thrown? Use the value \( g = 9.800 \, \text{m/s}^2 \) for the free-fall acceleration. --- **PROBLEM-SOLVING STRATEGY 4.1: Projectile motion problems** **MODEL:** Is it reasonable to ignore air resistance? If so, use the projectile motion model. **VISUALIZE:** Establish a coordinate system with the x-axis horizontal and the y-axis vertical. Define symbols and identify what the problem is trying to find. For a launch at angle \( \theta \), the initial velocity components are \( v_{ix} = v \cos \theta \) and \( v_{iy} = v \sin \theta \). **SOLVE:** The acceleration is known: \( a_x = 0 \) and \( a_y = -g \). Thus, the problem becomes one of two-dimensional kinematics. The kinematic equations are \[ \begin{array}{lc} \text{Horizontal} & \text{Vertical} \\ x_f = x_i + v_{ix} \Delta t & y_f = y_i + v_{iy} \Delta t - \frac{1}{2} g (\Delta t)^2 \, , \\ v_{tx} = v_{ix} = \text{constant} & v_{ty} = v_{iy} - g \Delta t \, \end{array} \] \( \Delta t \) is the same for the horizontal and vertical components of the motion. Find \( \Delta t \) from one component, and then use that value for the other component. **REVIEW:** Check that your result has the correct units and significant figures, is reasonable, and answers the question. --- **Model** Start by making simplifying assumptions: Model the rock as a particle in free fall. You can ignore air resistance because the rock is a relatively heavy object moving relatively slowly.
### Question: Which diagram represents an accurate sketch of the rock's trajectory? ### Diagrams Explanation: The provided diagrams illustrate the trajectory of a rock thrown into the air, showcasing different combinations of initial velocity (\( \vec{v_i} \)), acceleration (\( \vec{a} \)), and final velocity (\( \vec{v_f} \)). Below is a detailed description of each diagram: 1. **First Diagram:** - **Initial Velocity (\( \vec{v_i} \))**: The vector is directed upwards and to the right. - **Acceleration (\( \vec{a} \))**: An orange vector pointed straight downwards. - **Final Velocity (\( \vec{v_f} \))**: The vector is directed downwards. - The trajectory shown is a parabolic arc with the rock returning to the ground. 2. **Second Diagram:** - **Initial Velocity (\( \vec{v_i} \))**: The vector is directed to the left. - **Acceleration (\( \vec{a} \))**: An orange vector pointed straight downwards. - **Final Velocity (\( \vec{v_f} \))**: The vector is directed downwards. - The trajectory forms a parabolic arc, starting from left to right, and descending to the ground. 3. **Third Diagram:** - **Initial Velocity (\( \vec{v_i} \))**: The vector is directed upwards and slightly right. - **Acceleration (\( \vec{a} \))**: An orange vector pointed straight downwards. - **Final Velocity (\( \vec{v_f} \))**: The vector is directed diagonally downwards to the right. - The trajectory is a symmetric parabolic arc with the rock landing on the right. 4. **Fourth Diagram:** - **Initial Velocity (\( \vec{v_i} \))**: The vector is directed upwards and to the left. - **Acceleration (\( \vec{a} \))**: An orange vector pointed upwards. - **Final Velocity (\( \vec{v_f} \))**: The vector is directed diagonally downwards to the right. - The trajectory forms an upward arc to the left and descends to the ground on the right. For an educational focus, these diagrams are used to
Transcribed Image Text:### Question: Which diagram represents an accurate sketch of the rock's trajectory? ### Diagrams Explanation: The provided diagrams illustrate the trajectory of a rock thrown into the air, showcasing different combinations of initial velocity (\( \vec{v_i} \)), acceleration (\( \vec{a} \)), and final velocity (\( \vec{v_f} \)). Below is a detailed description of each diagram: 1. **First Diagram:** - **Initial Velocity (\( \vec{v_i} \))**: The vector is directed upwards and to the right. - **Acceleration (\( \vec{a} \))**: An orange vector pointed straight downwards. - **Final Velocity (\( \vec{v_f} \))**: The vector is directed downwards. - The trajectory shown is a parabolic arc with the rock returning to the ground. 2. **Second Diagram:** - **Initial Velocity (\( \vec{v_i} \))**: The vector is directed to the left. - **Acceleration (\( \vec{a} \))**: An orange vector pointed straight downwards. - **Final Velocity (\( \vec{v_f} \))**: The vector is directed downwards. - The trajectory forms a parabolic arc, starting from left to right, and descending to the ground. 3. **Third Diagram:** - **Initial Velocity (\( \vec{v_i} \))**: The vector is directed upwards and slightly right. - **Acceleration (\( \vec{a} \))**: An orange vector pointed straight downwards. - **Final Velocity (\( \vec{v_f} \))**: The vector is directed diagonally downwards to the right. - The trajectory is a symmetric parabolic arc with the rock landing on the right. 4. **Fourth Diagram:** - **Initial Velocity (\( \vec{v_i} \))**: The vector is directed upwards and to the left. - **Acceleration (\( \vec{a} \))**: An orange vector pointed upwards. - **Final Velocity (\( \vec{v_f} \))**: The vector is directed diagonally downwards to the right. - The trajectory forms an upward arc to the left and descends to the ground on the right. For an educational focus, these diagrams are used to
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Part B As stated in the strategy, choose a coordinate system where the x axis is horizontal and the y axis is vertical. Note that in the strategy, the y component of the projectile's acceleration, ay, is taken to be negative. This implies that the positive y axis is upward. Use the same convention for your y axis, and take the positive x axis to be to the right. Where you choose your origin doesn't change the answer to the question, but choosing an origin can make a problem easier to solve (even if only a bit). Usually it is nice if the majority of the quantities you are given and the quantity you are trying to solve for take positive values relative to your chosen origin. Given this goal, what location for the origin of the coordinate system would make this problem easiest? At the peak of the trajectory At the point where the rock strikes the ground At the point where the rock is released At ground level below the point where the rock is launched At ground level below the peak of the trajectory Submit Request Answer
Transcribed Image Text:Part B As stated in the strategy, choose a coordinate system where the x axis is horizontal and the y axis is vertical. Note that in the strategy, the y component of the projectile's acceleration, ay, is taken to be negative. This implies that the positive y axis is upward. Use the same convention for your y axis, and take the positive x axis to be to the right. Where you choose your origin doesn't change the answer to the question, but choosing an origin can make a problem easier to solve (even if only a bit). Usually it is nice if the majority of the quantities you are given and the quantity you are trying to solve for take positive values relative to your chosen origin. Given this goal, what location for the origin of the coordinate system would make this problem easiest? At the peak of the trajectory At the point where the rock strikes the ground At the point where the rock is released At ground level below the point where the rock is launched At ground level below the peak of the trajectory Submit Request Answer
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