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. `(hu^(2))/(gb^(2))`
B. `(u^(2)g)/(gb^(2))`
C. `(2hu^(2))/(gb^(2))`
D. `(2u^(2)g)/(hb^(2))`

Explanation

To solve the problem of a ball rolling off the top of a staircase and hitting the edge of the nth step, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The ball rolls off the top of the staircase with a horizontal velocity `u m/s`.
- Each step is `h` meters high and `b` meters wide.
- We need to determine the condition under which the ball hits the edge of the nth step.

2. **Identifying Distances**:
- If the ball hits the nth step, the horizontal distance it travels is `nb` (n steps wide) and the vertical distance it falls is `nh` (n steps high).

3. **Using the Projectile Motion Equation**:
- The vertical motion of the ball can be described by the equation of motion for a projectile:
\[ y = \frac{g x^2}{2u^2} \]
- Here, `y` is the vertical distance fallen (which is `nh`), and `x` is the horizontal distance traveled (which is `nb`).

4. **Substituting the Distances**:
- Substitute `y` with `nh` and `x` with `nb` in the projectile motion equation:
\[ nh = \frac{g (nb)^2}{2u^2} \]

5. **Simplifying the Equation**:
- Rearranging the equation gives:
\[ nh = \frac{g n^2 b^2}{2u^2} \]
- We can cancel `n` from both sides (assuming `n ≠ 0`):
\[ h = \frac{g n b^2}{2u^2} \]

6. **Solving for n**:
- Rearranging the equation to solve for `n` gives:
\[ n = \frac{2uh}{gb^2} \]

7. **Conclusion**:
- The ball will hit the edge of the nth step if:
\[ n = \frac{2uh}{gb^2} \]