bottom! 1. A ball is dropped from a height of 1 m and loses 10 percent of its kinetic energy when it bounces on the ground. To what height does it then rise? alm is struck by a

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**Physics Problem: Ball Drop and Energy Loss**

**Problem Statement:**
A ball is dropped from a height of 1 meter and loses 10 percent of its kinetic energy when it bounces on the ground. To what height does it then rise?

---

**Solution Approach:**

1. **Initial Potential Energy (PE):**
   - The ball is dropped from a height of \( h = 1 \) meter.
   - Potential Energy at height \( h \) is given by \( PE = mgh \), where \( m \) is the mass of the ball, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)), and \( h \) is the height.

2. **Conversion to Kinetic Energy:**
   - When the ball is at its maximum height, the potential energy is fully converted to kinetic energy just before it hits the ground (ignoring air resistance).
   - Therefore, the kinetic energy \( KE \) just before impact is equal to the initial potential energy, i.e., \( KE = mgh \).

3. **Energy Loss on Impact:**
   - The ball loses 10 percent of its kinetic energy upon impact.
   - Remaining kinetic energy after the bounce = \( 0.9 \times KE \).

4. **Conversion Back to Potential Energy:**
   - After the bounce, the remaining kinetic energy is converted back into potential energy at the new height \( h_{\text{new}} \).
   - Therefore, \( 0.9 \times mgh = mgh_{\text{new}} \).

5. **Solving for New Height \( h_{\text{new}} \):**
   - Cancel the mass \( m \) from both sides of the equation:
     \[
     0.9 \times gh = gh_{\text{new}}
     \]
   - Simplify to find:
     \[
     h_{\text{new}} = 0.9 \times h
     \]
   - Since \( h = 1 \) meter initially:
     \[
     h_{\text{new}} = 0.9 \times 1 \text{ meter} = 0.9 \text{ meters}
     \]

**Therefore, the ball rises to a height of 0.9 meters after bouncing.**
Transcribed Image Text:**Physics Problem: Ball Drop and Energy Loss** **Problem Statement:** A ball is dropped from a height of 1 meter and loses 10 percent of its kinetic energy when it bounces on the ground. To what height does it then rise? --- **Solution Approach:** 1. **Initial Potential Energy (PE):** - The ball is dropped from a height of \( h = 1 \) meter. - Potential Energy at height \( h \) is given by \( PE = mgh \), where \( m \) is the mass of the ball, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)), and \( h \) is the height. 2. **Conversion to Kinetic Energy:** - When the ball is at its maximum height, the potential energy is fully converted to kinetic energy just before it hits the ground (ignoring air resistance). - Therefore, the kinetic energy \( KE \) just before impact is equal to the initial potential energy, i.e., \( KE = mgh \). 3. **Energy Loss on Impact:** - The ball loses 10 percent of its kinetic energy upon impact. - Remaining kinetic energy after the bounce = \( 0.9 \times KE \). 4. **Conversion Back to Potential Energy:** - After the bounce, the remaining kinetic energy is converted back into potential energy at the new height \( h_{\text{new}} \). - Therefore, \( 0.9 \times mgh = mgh_{\text{new}} \). 5. **Solving for New Height \( h_{\text{new}} \):** - Cancel the mass \( m \) from both sides of the equation: \[ 0.9 \times gh = gh_{\text{new}} \] - Simplify to find: \[ h_{\text{new}} = 0.9 \times h \] - Since \( h = 1 \) meter initially: \[ h_{\text{new}} = 0.9 \times 1 \text{ meter} = 0.9 \text{ meters} \] **Therefore, the ball rises to a height of 0.9 meters after bouncing.**
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