A steel ball is dropped on a hard surface from a height of `1m` and rebounds to a height of `64cm`. The maximum height attained by the ball after `n^(th)` bounce is (in m)

Options

A. (a) `(0.64)^(2n)`
B. (b) `(0.8)^(2n)`
C. (c) `(0.5)^(2n)`
D. (d) `(0.8)^n`

Explanation

To solve the problem, we need to find the maximum height attained by the steel ball after the nth bounce. The ball is dropped from a height of 1 meter and rebounds to a height of 64 cm after the first bounce.

### Step-by-Step Solution:

1. **Identify the Initial Height and Rebound Height**:
- The initial height (H) from which the ball is dropped is 1 meter (100 cm).
- The height to which the ball rebounds after the first bounce (h) is 64 cm.

2. **Calculate the Coefficient of Restitution (e)**:
- The coefficient of restitution (e) is defined as the ratio of the rebound height to the original height.
- We can express this as:
\[ e = \sqrt{\frac{h}{H}} = \sqrt{\frac{64 \text{ cm}}{100 \text{ cm}}} \]
- Simplifying this gives:
\[ e = \sqrt{0.64} = 0.8 \]

3. **Determine the Height After nth Bounce**:
- The height attained after the nth bounce can be calculated using the formula:
\[ h_n = e^{2n} \cdot H \]
- Substituting the values we have:
\[ h_n = (0.8)^{2n} \cdot 1 \text{ m} \]
- This simplifies to:
\[ h_n = 0.8^{2n} \text{ m} \]

4. **Final Expression**:
- Thus, the maximum height attained by the ball after the nth bounce is:
\[ h_n = 0.8^{2n} \text{ m} \]

### Conclusion:
The maximum height attained by the ball after the nth bounce is \( 0.8^{2n} \) meters.