A staircase contains three steps each 10 cm high and 20 cm wide. What should be the minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane

The correct answer is 2 m/s.

Let us analyze the motion of the ball as a projectile. The ball rolls off the uppermost plane horizontally with a velocity $u$ from the edge, which we can take as the origin $(0,0)$.

The staircase has three steps, each with a height of $10\text{cm}=0.1\text{m}$ and a width of $20\text{cm}=0.2\text{m}$. Let us establish the coordinates of the outer edges of the steps:
- Edge of the first step: $(0.2\text{m},-0.1\text{m})$
- Edge of the second step: $(0.4\text{m},-0.2\text{m})$
- The lowest plane (third step) starts beyond $x=0.4\text{m}$ at a depth of $y=-0.3\text{m}$.

For the ball to hit the lowest plane directly, its trajectory must clear the edge of the second step. This means that at a horizontal distance of $x=0.4\text{m}$, the vertical downward displacement of the ball must be less than $0.2\text{m}$.

The horizontal distance traveled in time $t$ is given by:
$x=ut\Rightarrow t=\frac{x}{u}$

The vertical downward distance traveled in time $t$ under gravity is given by:
$y=\frac{1}{2}g{t}^2=\frac{1}{2}g{\left(\frac{x}{u}\right)}^2$

Using $g=10{\text{m/s}}^2$, we substitute the coordinates of the second step's edge $x=0.4\text{m}$ and require the vertical fall $y$ to be less than $0.2\text{m}$ at this point:
$\frac{1}{2}\cdot 10\cdot {\left(\frac{0.4}{u}\right)}^2<0.2$

Simplifying the inequality:
$5\cdot \frac{0.16}{u^2}$ < 0.2
$\frac{0.8}{u^2}$ < 0.2
$u^2>\frac{0.8}{0.2}$
$u^2>4$
$u>2\text{m/s}$

Therefore, the minimum horizontal velocity required is 2 m/s.