A ball rolls off top of a staircase with a horizontal velocity u m/s. If the steps are h metre high and b mere wide, the ball will just hit the edge of nth step if n equals to

The correct option is 2hu²/gb².

To find the step number $n$ at which the ball will hit the edge, we can analyze the motion of the ball as a projectile projected horizontally from the top of the staircase.

Let the top of the staircase be the origin of our coordinate system $(0,0)$.

Each step has a height of $h$ meters and a width of $b$ meters. For the ball to just hit the edge of the $n$-th step, its horizontal and vertical coordinates must satisfy:
Horizontal distance, $x=nb$
Vertical distance (downwards), $y=nh$

Now, let's consider the horizontal motion of the ball. Since there is no acceleration in the horizontal direction, the horizontal distance traveled in time $t$ is given by:
$x=ut$
Substituting $x=nb$ into this equation:
$nb=ut\Rightarrow t=\frac{nb}{u}$

Next, we analyze the vertical motion of the ball. Since the initial vertical velocity is zero and the acceleration is gravity $g$ acting downwards, the vertical distance covered in time $t$ is given by:
$y=\frac{1}{2}g{t}^2$

Substituting $y=nh$ and the value of $t$ from the horizontal motion:
$nh=\frac{1}{2}g{\left(\frac{nb}{u}\right)}^2$
Simplifying this equation:
$nh=\frac{g{n}^2{b}^2}{2u^2}$

Since $n\ne 0$, we can divide both sides by $n$ to solve for $n$:
$h=\frac{gn{b}^2}{2u^2}$
Rearranging the formula to isolate $n$ gives:
$n=\frac{2h{u}^2}{g{b}^2}$