3. A stone is released from rest and dropped into a deep well. Eight seconds later, the sound of the stone splashing into the water at the bottom of the well returns to the ear of the person who released the stone. How long does it take the stone to drop to the bottom of the well? How deep is the well? Ignore air resistance. Note: The speed of sound is 340 m/s.
3. A stone is released from rest and dropped into a deep well. Eight seconds later, the sound of the stone splashing into the water at the bottom of the well returns to the ear of the person who released the stone. How long does it take the stone to drop to the bottom of the well? How deep is the well? Ignore air resistance. Note: The speed of sound is 340 m/s.
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Please see question and answer from the student manual attached. Can you please go over step-by-step how to arrive at the solutions?
![### Physics Problem
**Problem Statement:**
A stone is released from rest and dropped into a deep well. Eight seconds later, the sound of the stone splashing into the water at the bottom of the well returns to the ear of the person who released the stone.
**Questions:**
1. How long does it take the stone to drop to the bottom of the well?
2. How deep is the well?
*Note*: Ignore air resistance. The speed of sound is 340 m/s.
---
**Solution:**
**Step-by-Step Explanation:**
**1. Identifying the equations:**
The depth of the well \( d \) is determined by the equation of motion:
\[ d = 4.9t^2 \]
where \( t \) is the time it takes the stone to hit the water.
The sound of the splash travels back to the top of the well after the stone hits the water. The total time for the event is 8 seconds. If \( t \) is the time for the stone to fall, then \( (8 - t) \) is the time for the sound to travel back to the top.
Given the speed of sound \( v = 340 \, \text{m/s} \), the distance the sound travels is:
\[ d = 340(8 - t) \]
**2. Setting up the equation:**
Since both expressions for \( d \) represent the same depth, we equate them:
\[ 4.9t^2 = 340(8 - t) \]
**3. Solving the quadratic equation:**
\[ 4.9t^2 + 340t - 2720 = 0 \]
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4.9 \), \( b = -340 \), and \( c = -2720 \):
\[ t = \frac{340 \pm \sqrt{340^2 - 4 \cdot 4.9 \cdot (-2720)}}{2 \cdot 4.9} \]
\[ t = \frac{340 \pm \sqrt{115600 + 53192}}{9.8} \]
\[ t = \frac{340 \pm \sqrt{168792}}{9.8} \]
\[ t](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1b1372ae-e668-4971-89ee-29da7ac7c466%2F33201b78-5211-41b1-a36d-5cc583bdf13f%2Fvb2qkxh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Physics Problem
**Problem Statement:**
A stone is released from rest and dropped into a deep well. Eight seconds later, the sound of the stone splashing into the water at the bottom of the well returns to the ear of the person who released the stone.
**Questions:**
1. How long does it take the stone to drop to the bottom of the well?
2. How deep is the well?
*Note*: Ignore air resistance. The speed of sound is 340 m/s.
---
**Solution:**
**Step-by-Step Explanation:**
**1. Identifying the equations:**
The depth of the well \( d \) is determined by the equation of motion:
\[ d = 4.9t^2 \]
where \( t \) is the time it takes the stone to hit the water.
The sound of the splash travels back to the top of the well after the stone hits the water. The total time for the event is 8 seconds. If \( t \) is the time for the stone to fall, then \( (8 - t) \) is the time for the sound to travel back to the top.
Given the speed of sound \( v = 340 \, \text{m/s} \), the distance the sound travels is:
\[ d = 340(8 - t) \]
**2. Setting up the equation:**
Since both expressions for \( d \) represent the same depth, we equate them:
\[ 4.9t^2 = 340(8 - t) \]
**3. Solving the quadratic equation:**
\[ 4.9t^2 + 340t - 2720 = 0 \]
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4.9 \), \( b = -340 \), and \( c = -2720 \):
\[ t = \frac{340 \pm \sqrt{340^2 - 4 \cdot 4.9 \cdot (-2720)}}{2 \cdot 4.9} \]
\[ t = \frac{340 \pm \sqrt{115600 + 53192}}{9.8} \]
\[ t = \frac{340 \pm \sqrt{168792}}{9.8} \]
\[ t
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