1. Set up the projectile motion

Take the top edge of the stairway as origin.

• Horizontal direction: $x$
• Vertical downward direction: $y$

The ball is projected horizontally with speed $u$.

So its motion is: $x = ut$ $y = \frac{1}{2}gt^2$

Eliminating $t$: $y = \frac{g x^2}{2u^2}$

Given $g=10\,\text{m/s}^2$, $y = \frac{5x^2}{u^2}$

2. Equation of the 5th step

Each step has:

• height $h = 0.1\,\text{m}$
• width $b = 0.1\,\text{m}$

So the horizontal surface of the 5th step lies at vertical depth $y = 5h = 0.5\,\text{m}$ and extends horizontally from $x = 4b = 0.4\,\text{m} \quad \text{to} \quad x = 5b = 0.5\,\text{m}.$

For the ball to just hit step 5 with minimum velocity, its trajectory should pass through the outer edge of the 5th step, i.e. the point $(x,y)=(0.4,0.5).$

If the speed were any smaller, it would hit an earlier step. This gives the minimum speed to first reach step 5.

3. Substitute into projectile equation

Using $y = \frac{5x^2}{u^2},$ at $(x,y)=(0.4,0.5)$:

$0.5 = \frac{5(0.4)^2}{u^2}$

$0.5 = \frac{5\times 0.16}{u^2}$

$0.5 = \frac{0.8}{u^2}$

$u^2 = \frac{0.8}{0.5} = 1.6$

Thus, $u = \sqrt{1.6} = \sqrt{\frac{8}{5}}\,\text{m/s}$

The question says $u=\sqrt{x}\,\text{m/s}$, so $x=1.6=\frac{8}{5}.$

Since the answer is required as an integer, we interpret it as $10x = 16$ only if scaled, but as written, $x$ itself is not an integer.

However, in standard JEE formulation, one usually writes $u = \sqrt{\frac{x}{5}}$ or similar. Here the direct value from the given wording is: $x=1.6$

4. Check against stored answer

Stored correct answer: $2$

My derived value is: $x=1.6$ not $2$.

So I do not agree with the stored answer. The likely issue is either:

• the wording of the problem, or
• the stored answer being rounded incorrectly.

If rounded to nearest integer, $x\approx 2$, but that is not exact.