A metal ball falls from a height of 32 metre on a steel plate. If the coefficient of restitution is 0.5, to what height will the ball rise after second bounce

Options

A. 2 m
B. 4m
C. 8m
D. 16 m

Explanation

To solve the problem of determining the height to which a metal ball will rise after the second bounce, given that it falls from a height of 32 meters and has a coefficient of restitution of 0.5, we can follow these steps:

### Step-by-Step Solution:

1. **Initial Height and Potential Energy**:
- The ball is dropped from a height \( h = 32 \) m.
- The potential energy (PE) at this height is given by:
\[ PE = mgh \]
- Here, \( m \) is the mass of the ball and \( g \) is the acceleration due to gravity.

2. **Velocity Just Before Impact**:
- Just before hitting the ground, all potential energy converts to kinetic energy (KE):
\[ KE = \frac{1}{2} mv^2 = mgh \]
- Canceling \( m \) from both sides, we find the velocity \( v \) just before impact:
\[ v = \sqrt{2gh} \]

3. **Velocity After First Bounce**:
- The coefficient of restitution \( e \) is defined as:
\[ e = \frac{v'}{v} \]
- Given \( e = 0.5 \), we can express the upward velocity \( v' \) after the first bounce:
\[ v' = e \cdot v = 0.5 \cdot \sqrt{2gh} \]

4. **Height After First Bounce**:
- The kinetic energy after the first bounce converts back to potential energy at the new height \( h' \):
\[ \frac{1}{2} mv'^2 = mgh' \]
- Canceling \( m \) and substituting \( v' \):
\[ h' = \frac{(0.5 \cdot \sqrt{2gh})^2}{2g} \]
- Simplifying:
\[ h' = \frac{0.25 \cdot 2gh}{2g} = 0.25h = 0.5^2 \cdot h \]
- Thus, the height after the first bounce is:
\[ h' = 0.25 \cdot 32 = 8 \text{ meters} \]

5. **Height After Second Bounce**:
- For the second bounce, we apply the same logic:
\[ h'' = 0.5^2 \cdot h' = 0.5^2 \cdot (0.5^2 \cdot h) = 0.5^4 \cdot h \]
- Therefore, substituting \( h = 32 \):
\[ h'' = 0.5^4 \cdot 32 = \frac{32}{16} = 2 \text{ meters} \]

### Final Answer:
The height to which the ball will rise after the second bounce is **2 meters**.