A ball rolls off the top of a stairway with a horizontal velocity `u`. If the height and width of steps are `b` and `h` respectively and the ball hits the edge of `n^(th)` step, then `n` is equal to

Options

A. `(hu^(2))/(gb^(2))`
B. `(hu^(2))/(2 g b^(2))`
C. `(2h u^(2))/(gb^(2))`
D. `(hu^(2))/(4 gb^(2))`

Explanation

To solve the problem, we need to analyze the motion of the ball as it rolls off the stairway and hits the edge of the \( n^{th} \) step. We will use the concepts of projectile motion and the geometry of the steps.

### Step-by-Step Solution:

1. **Understanding the Geometry of the Steps**:
- Each step has a height \( h \) and a width \( b \).
- If the ball hits the \( n^{th} \) step, the vertical distance it falls is \( n \cdot h \) (since it falls \( h \) for each step).

2. **Setting Up the Motion Equations**:
- The ball rolls off horizontally with an initial velocity \( u \).
- The horizontal distance traveled when it hits the \( n^{th} \) step is \( n \cdot b \) (since it moves \( b \) for each step).

3. **Time of Flight**:
- The time \( t \) taken to fall a vertical distance \( n \cdot h \) can be calculated using the equation of motion under gravity:
\[ h = \frac{1}{2} g t^2 \]
Rearranging gives:
\[ t = \sqrt{\frac{2n h}{g}} \]

4. **Horizontal Motion**:
- The horizontal distance \( x \) covered in time \( t \) is given by:
\[ x = u \cdot t \]
- Substituting the expression for \( t \):
\[ n \cdot b = u \cdot \sqrt{\frac{2n h}{g}} \]

5. **Squaring Both Sides**:
- To eliminate the square root, we square both sides:
\[ (n \cdot b)^2 = u^2 \cdot \frac{2n h}{g} \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ n^2 b^2 = \frac{2n h u^2}{g} \]
- Dividing both sides by \( n \) (assuming \( n \neq 0 \)):
\[ n b^2 = \frac{2h u^2}{g} \]

7. **Solving for \( n \)**:
- Finally, we can solve for \( n \):
\[ n = \frac{2h u^2}{g b^2} \]

### Final Result:
The value of \( n \) is given by:
\[ n = \frac{2h u^2}{g b^2} \]