The Figure below shows the speed v versus height y of a ball tossed directly upward, along a y-axis. Distance d is 0.40 m. The speed at height ya is vA. The speed at height yg is VA. VWhat is VA speed va in terms of the given quantities? (Use g 10 m/s?.)
The Figure below shows the speed v versus height y of a ball tossed directly upward, along a y-axis. Distance d is 0.40 m. The speed at height ya is vA. The speed at height yg is VA. VWhat is VA speed va in terms of the given quantities? (Use g 10 m/s?.)
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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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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![### Example Problem: Speed vs. Height of a Tossed Ball
The figure below shows the speed \( v \) versus height \( y \) of a ball tossed directly upward along a y-axis.
- Distance \( d \) is 0.40 m.
- The speed at height \( y_A \) is \( v_A \).
- The speed at height \( y_B \) is \( \frac{1}{3} v_A \).
**Question:** What is speed \( v_A \) in terms of the given quantities?
**Given:** Use \( g = 10 \, \text{m/s}^2 \).
#### Diagram Explanation:
The graph displays the relationship between speed (\( v \)) and height (\( y \)) for the ball. Key points to note from the graph:
1. The vertical axis (y-axis) represents speed \( v \).
2. The horizontal axis (x-axis) represents height \( y \).
3. There are horizontal dashed lines indicating speeds \( v_A \) and \( \frac{1}{3} v_A \).
4. Heights \( y_A \) and \( y_B \) are marked on the \( y \)-axis with \( d \) as the distance between \( y_A \) and \( y_B \).
The graph suggests a downward-opening parabolic relationship between \( v \) and \( y \).
The goal is to determine the speed \( v_A \) using the provided information and the given acceleration due to gravity \( g \).
#### Detailed Solution:
To solve for \( v_A \), we analyze the energy changes when the ball moves from height \( y_A \) to \( y_B \).
1. **Kinetic Energy at \( y_A \):**
\[
K.E. = \frac{1}{2} m v_A^2
\]
2. **Kinetic Energy at \( y_B \):**
\[
K.E. = \frac{1}{2} m \left( \frac{1}{3} v_A \right)^2 = \frac{1}{2} m \left( \frac{1}{9} v_A^2 \right) = \frac{1}{18} m v_A^2
\]
3. **Potential Energy Change:**
\[
\Delta P.E. = m g d = m \times](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffd14f466-2175-4b26-80f9-e0672840c669%2F31bf890e-2063-4f3c-9e79-f85312d40fda%2Fm6fpjp_processed.png&w=3840&q=75)
Transcribed Image Text:### Example Problem: Speed vs. Height of a Tossed Ball
The figure below shows the speed \( v \) versus height \( y \) of a ball tossed directly upward along a y-axis.
- Distance \( d \) is 0.40 m.
- The speed at height \( y_A \) is \( v_A \).
- The speed at height \( y_B \) is \( \frac{1}{3} v_A \).
**Question:** What is speed \( v_A \) in terms of the given quantities?
**Given:** Use \( g = 10 \, \text{m/s}^2 \).
#### Diagram Explanation:
The graph displays the relationship between speed (\( v \)) and height (\( y \)) for the ball. Key points to note from the graph:
1. The vertical axis (y-axis) represents speed \( v \).
2. The horizontal axis (x-axis) represents height \( y \).
3. There are horizontal dashed lines indicating speeds \( v_A \) and \( \frac{1}{3} v_A \).
4. Heights \( y_A \) and \( y_B \) are marked on the \( y \)-axis with \( d \) as the distance between \( y_A \) and \( y_B \).
The graph suggests a downward-opening parabolic relationship between \( v \) and \( y \).
The goal is to determine the speed \( v_A \) using the provided information and the given acceleration due to gravity \( g \).
#### Detailed Solution:
To solve for \( v_A \), we analyze the energy changes when the ball moves from height \( y_A \) to \( y_B \).
1. **Kinetic Energy at \( y_A \):**
\[
K.E. = \frac{1}{2} m v_A^2
\]
2. **Kinetic Energy at \( y_B \):**
\[
K.E. = \frac{1}{2} m \left( \frac{1}{3} v_A \right)^2 = \frac{1}{2} m \left( \frac{1}{9} v_A^2 \right) = \frac{1}{18} m v_A^2
\]
3. **Potential Energy Change:**
\[
\Delta P.E. = m g d = m \times
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