A ball rolls off the top of a staircase with a horizontal velocity `u m//s`. If the steps are h meter high and b meter wide, the ball will hit the edge of the nth steps, if:

Options

A. `n = (2 h u)/(g b^2)`
B. `n = (2 h u^2)/(g b)`
C. `n = (2 h u^2)/(g b^2)`
D. `n = (h u^2)/(g b^2)`

Explanation

To solve the problem, we need to analyze the motion of the ball as it rolls off the staircase. The ball has a horizontal velocity \( u \) and the staircase has steps that are \( h \) meters high and \( b \) meters wide. We want to find out how many steps \( n \) the ball will hit.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The ball rolls off horizontally, so it has an initial horizontal velocity \( u \).
- The vertical motion is influenced by gravity, which accelerates the ball downwards.

2. **Horizontal Motion**:
- The horizontal distance traveled by the ball when it hits the edge of the \( n \)-th step can be expressed as:
\[ \text{Horizontal Distance} = u \cdot t \]
- Since the width of each step is \( b \), the total horizontal distance to the \( n \)-th step is:
\[ \text{Horizontal Distance} = n \cdot b \]
- Therefore, we can equate the two expressions:
\[ n \cdot b = u \cdot t \]

3. **Vertical Motion**:
- The vertical distance fallen by the ball when it hits the \( n \)-th step is the height of \( n \) steps, which is:
\[ \text{Vertical Distance} = n \cdot h \]
- The vertical motion can be described by the second equation of motion:
\[ \text{Vertical Distance} = \frac{1}{2} g t^2 \]
- Equating the two expressions for vertical distance gives:
\[ n \cdot h = \frac{1}{2} g t^2 \]

4. **Finding Time \( t \)**:
- From the horizontal motion equation, we can express time \( t \):
\[ t = \frac{n \cdot b}{u} \]

5. **Substituting \( t \) into the Vertical Motion Equation**:
- Substitute \( t \) into the vertical motion equation:
\[ n \cdot h = \frac{1}{2} g \left(\frac{n \cdot b}{u}\right)^2 \]
- This simplifies to:
\[ n \cdot h = \frac{1}{2} g \cdot \frac{n^2 \cdot b^2}{u^2} \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ 2h \cdot u^2 = g \cdot n \cdot b^2 \]
- Solving for \( n \):
\[ n = \frac{2h \cdot u^2}{g \cdot b^2} \]

### Final Answer:
The ball will hit the edge of the \( n \)-th step if:
\[ n = \frac{2h \cdot u^2}{g \cdot b^2} \]