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

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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
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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