A `70 g` ball B dropped from a height `h_(0) = 9m` reaches a height `h_(2) = 0.25 m` after bouncing twice from identical `210 g` plates. Plate A rests directly on hard ground, while plate C rests on a foam-rubber mat. Determine
(i) the coefficient of restituation between the ball and the plates,
(ii) the height `h_(1)` of the ball's first bounce.

Explanation

To solve the problem, we will break it down into two parts as requested:

### Part (i): Determine the coefficient of restitution (e)

1. **Identify the initial conditions**:
- Mass of the ball, \( m_B = 70 \, \text{g} = 0.07 \, \text{kg} \)
- Initial height, \( h_0 = 9 \, \text{m} \)
- Height after two bounces, \( h_2 = 0.25 \, \text{m} \)
- Mass of the plates, \( m_A = m_C = 210 \, \text{g} = 0.21 \, \text{kg} \)

2. **Calculate the initial velocity of the ball just before the first impact**:
- Using the equation of motion, the velocity \( v_0 \) just before impact can be calculated using:
\[ v_0 = \sqrt{2gh_0} \]
- Substituting \( g = 9.81 \, \text{m/s}^2 \) and \( h_0 = 9 \, \text{m} \):
\[ v_0 = \sqrt{2 \times 9.81 \times 9} = \sqrt{176.58} \approx 13.29 \, \text{m/s} \]

3. **Apply conservation of momentum for the first collision**:
- Let \( v_1 \) be the velocity of the ball after the first bounce. The coefficient of restitution \( e \) is defined as:
\[ e = \frac{\text{relative velocity after collision}}{\text{relative velocity before collision}} \]
- The relative velocity before collision is \( v_0 \) (the ball is moving downwards) and after collision is \( -v_1 \) (the ball is moving upwards):
\[ e = \frac{-v_1}{v_0} \]

4. **Determine the velocity after the first bounce**:
- For the second collision, the ball collides with plate C. The momentum conservation gives:
\[ m_B v_1 = (m_B + m_C) v' \]
- Here, \( v' \) is the velocity of the combined system after the second collision. The velocity of the ball after the second collision can be expressed as:
\[ v' = \frac{m_B v_1}{m_B + m_C} \]

5. **Using the second bounce height**:
- The height after the second bounce \( h_2 \) is given by:
\[ h_2 = e^2 h_0 \]
- Rearranging gives:
\[ e^2 = \frac{h_2}{h_0} = \frac{0.25}{9} \]
- Thus:
\[ e = \sqrt{\frac{0.25}{9}} = \frac{0.5}{3} = \frac{1}{6} \approx 0.333 \]

### Part (ii): Determine the height \( h_1 \) of the ball's first bounce

1. **Using the coefficient of restitution**:
- The height after the first bounce \( h_1 \) can be calculated using:
\[ h_1 = e^2 h_0 \]

2. **Substituting the value of \( e \)**:
- From the previous calculation, we found \( e \approx 0.333 \):
\[ h_1 = (0.333)^2 \times 9 = \frac{1}{9} \times 9 = 1 \, \text{m} \]

### Final Answers:
- (i) The coefficient of restitution \( e \approx 0.333 \)
- (ii) The height \( h_1 \) of the ball's first bounce is \( 1 \, \text{m} \)