Alice shoots a pebble horizontally with her slingshot from a window that is at a height H and it lands a distance D away from the building. She then shoots a pebble with the same initial velocity from a different window, but this time the pebble lands a distance D/2 away. How high is the window from which Alice takes the second shot? A. 4H B. 2H C. H D. H/2 E. H/4
Alice shoots a pebble horizontally with her slingshot from a window that is at a height H and it lands a distance D away from the building. She then shoots a pebble with the same initial velocity from a different window, but this time the pebble lands a distance D/2 away. How high is the window from which Alice takes the second shot? A. 4H B. 2H C. H D. H/2 E. H/4
College Physics
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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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![**Physics Problem: Projectile Motion**
**Problem Statement:**
Alice shoots a pebble horizontally with her slingshot from a window that is at a height \(H\) and it lands a distance \(D\) away from the building. She then shoots a pebble with the same initial velocity from a different window, but this time the pebble lands a distance \(D/2\) away. How high is the window from which Alice takes the second shot?
- A. \(4H\)
- B. \(2H\)
- C. \(H\)
- D. \(H/2\)
- E. \(H/4\) (Highlighted as the correct answer)
**Solution Explanation:**
To solve this problem, we need to understand the concepts of projectile motion and the relationship between the horizontal distance (range) and the height from which the object is projected.
### Step-by-Step Solution:
1. **Understanding Initial Conditions:**
- The first shot:
- Height: \( H \)
- Horizontal distance: \( D \)
2. **Second Shot:**
- Height: Unknown (let's denote it as \( h \))
- Horizontal distance: \( D / 2 \)
3. **Important Equations:**
- The horizontal distance (range) for a projectile shot with an initial horizontal velocity \( v \) and time of flight \( t \) is given by:
\[ R = vt \]
- The time of flight \( t \) of the projectile depends on the height \( h \) from which it is shot:
\[ t = \sqrt{\frac{2h}{g}} \]
Where \( g \) is the acceleration due to gravity.
4. **For the First Shot:**
- The time of flight from height \( H \):
\[ t_1 = \sqrt{\frac{2H}{g}} \]
- The horizontal range:
\[ D = v t_1 = v \sqrt{\frac{2H}{g}} \]
5. **For the Second Shot:**
- The time of flight from height \( h \):
\[ t_2 = \sqrt{\frac{2h}{g}} \]
- The horizontal range:
\[ \frac{D}{2} = v t_2 = v \sqrt{\frac{2h}{g](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc2c02f4b-0440-4781-818e-0767718c1837%2Fbd898558-8344-48f1-921f-8bff580dd003%2Fpw7y10l_processed.png&w=3840&q=75)
Transcribed Image Text:**Physics Problem: Projectile Motion**
**Problem Statement:**
Alice shoots a pebble horizontally with her slingshot from a window that is at a height \(H\) and it lands a distance \(D\) away from the building. She then shoots a pebble with the same initial velocity from a different window, but this time the pebble lands a distance \(D/2\) away. How high is the window from which Alice takes the second shot?
- A. \(4H\)
- B. \(2H\)
- C. \(H\)
- D. \(H/2\)
- E. \(H/4\) (Highlighted as the correct answer)
**Solution Explanation:**
To solve this problem, we need to understand the concepts of projectile motion and the relationship between the horizontal distance (range) and the height from which the object is projected.
### Step-by-Step Solution:
1. **Understanding Initial Conditions:**
- The first shot:
- Height: \( H \)
- Horizontal distance: \( D \)
2. **Second Shot:**
- Height: Unknown (let's denote it as \( h \))
- Horizontal distance: \( D / 2 \)
3. **Important Equations:**
- The horizontal distance (range) for a projectile shot with an initial horizontal velocity \( v \) and time of flight \( t \) is given by:
\[ R = vt \]
- The time of flight \( t \) of the projectile depends on the height \( h \) from which it is shot:
\[ t = \sqrt{\frac{2h}{g}} \]
Where \( g \) is the acceleration due to gravity.
4. **For the First Shot:**
- The time of flight from height \( H \):
\[ t_1 = \sqrt{\frac{2H}{g}} \]
- The horizontal range:
\[ D = v t_1 = v \sqrt{\frac{2H}{g}} \]
5. **For the Second Shot:**
- The time of flight from height \( h \):
\[ t_2 = \sqrt{\frac{2h}{g}} \]
- The horizontal range:
\[ \frac{D}{2} = v t_2 = v \sqrt{\frac{2h}{g
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