A small rolls of the top of a stairway horizontally with a velocity of `4.5 ms^(-1)`. Each step is 0.2 m high and 0.3 m wide. If g is `10ms^(-2)`, then the ball will strike the nth step where `n` is equal to (assume ball strike at the edge of the step).

Options

A. 10
B. 9
C. 8
D. 11

Explanation

To solve the problem, we need to determine the step number \( n \) where the ball strikes after rolling off the top of the stairway horizontally. The key points to consider are the horizontal and vertical motions of the ball.

### Step-by-Step Solution:

1. **Identify the parameters**:
- Horizontal velocity \( u = 4.5 \, \text{m/s} \)
- Height of each step \( h = 0.2 \, \text{m} \)
- Width of each step \( w = 0.3 \, \text{m} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

2. **Determine the time taken to fall to the nth step**:
- The height from which the ball falls to the nth step is given by:
\[ H = n \times h = n \times 0.2 \, \text{m} \]
- Using the equation of motion for vertical displacement:
\[ H = \frac{1}{2} g t^2 \]
- Substituting for \( H \):
\[ n \times 0.2 = \frac{1}{2} \times 10 \times t^2 \]
- Simplifying:
\[ n \times 0.2 = 5 t^2 \]
- Rearranging gives:
\[ t^2 = \frac{0.2n}{5} = 0.04n \]

3. **Determine the horizontal distance traveled in time \( t \)**:
- The horizontal distance \( R \) covered in time \( t \) is given by:
\[ R = u \times t \]
- Substituting \( u = 4.5 \, \text{m/s} \):
\[ R = 4.5 t \]

4. **Relate the horizontal distance to the number of steps**:
- The horizontal distance to the nth step is:
\[ R = n \times w = n \times 0.3 \, \text{m} \]
- Setting the two expressions for \( R \) equal:
\[ n \times 0.3 = 4.5 t \]

5. **Substituting \( t \)**:
- From \( t^2 = 0.04n \), we can find \( t \):
\[ t = \sqrt{0.04n} = 0.2\sqrt{n} \]
- Substituting \( t \) into the horizontal distance equation:
\[ n \times 0.3 = 4.5 \times (0.2\sqrt{n}) \]
- Simplifying:
\[ n \times 0.3 = 0.9\sqrt{n} \]

6. **Rearranging the equation**:
- Squaring both sides:
\[ (n \times 0.3)^2 = (0.9\sqrt{n})^2 \]
- This gives:
\[ 0.09n^2 = 0.81n \]
- Rearranging:
\[ 0.09n^2 - 0.81n = 0 \]
- Factoring out \( n \):
\[ n(0.09n - 0.81) = 0 \]
- Thus, \( n = 0 \) or \( 0.09n = 0.81 \):
\[ n = \frac{0.81}{0.09} = 9 \]

### Conclusion:
The ball will strike the 9th step.