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)...
Related questions
Question
![**Question 9:** How far has a ball fallen downward from rest by the time it is traveling 35 m/s downward?
*This problem involves understanding the concepts of free fall and the equations of motion. To solve this, we will assume the acceleration due to gravity is approximately 9.81 m/s² (downward) and there is no air resistance.*
**Solution Approach:**
1. **Use the Kinematic Equation:**
\[
v^2 = u^2 + 2as
\]
Where:
- \( v \) = final velocity (35 m/s)
- \( u \) = initial velocity (0 m/s, since it starts from rest)
- \( a \) = acceleration (9.81 m/s² downward due to gravity)
- \( s \) = distance fallen
2. **Substitute the known values:**
\[
35^2 = 0 + 2 \times 9.81 \times s
\]
3. **Solve for \( s \):**
\[
1225 = 19.62s
\]
\[
s = \frac{1225}{19.62}
\]
\[
s \approx 62.45 \text{ meters}
\]
Therefore, the ball has fallen approximately 62.45 meters downward by the time it is traveling at 35 m/s.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe44e0876-942a-4733-ac3b-899fc9beb500%2Fe41a6daf-1d9d-436f-a869-6ec02766395e%2Fyuw9ae_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 9:** How far has a ball fallen downward from rest by the time it is traveling 35 m/s downward?
*This problem involves understanding the concepts of free fall and the equations of motion. To solve this, we will assume the acceleration due to gravity is approximately 9.81 m/s² (downward) and there is no air resistance.*
**Solution Approach:**
1. **Use the Kinematic Equation:**
\[
v^2 = u^2 + 2as
\]
Where:
- \( v \) = final velocity (35 m/s)
- \( u \) = initial velocity (0 m/s, since it starts from rest)
- \( a \) = acceleration (9.81 m/s² downward due to gravity)
- \( s \) = distance fallen
2. **Substitute the known values:**
\[
35^2 = 0 + 2 \times 9.81 \times s
\]
3. **Solve for \( s \):**
\[
1225 = 19.62s
\]
\[
s = \frac{1225}{19.62}
\]
\[
s \approx 62.45 \text{ meters}
\]
Therefore, the ball has fallen approximately 62.45 meters downward by the time it is traveling at 35 m/s.
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