A staircase contains three steps, each 10 cm high and 20 cm wide,...

Let the edge of the uppermost plane be the origin $(0,0)$.

The staircase has steps of height $h = 10 \text{ cm} = 0.1 \text{ m}$ and width $w = 20 \text{ cm} = 0.2 \text{ m}$.

From the given figure, to directly hit the lowest plane, the ball must clear the edge of the second step.

The coordinates of the edge of the second step are $x = 2w = 0.4 \text{ m}$ and $y = -2h = -0.2 \text{ m}$.

The equation of the trajectory of the ball is given by:

$y = -\dfrac{gx^2}{2u^2}$

For the ball to just clear the second step, its trajectory must pass through the point $(0.4, -0.2)$. Substituting these values into the trajectory equation:

$-0.2 = -\dfrac{10 \times (0.4)^2}{2u^2}$

$0.2 = \dfrac{10 \times 0.16}{2u^2}$

$0.2 = \dfrac{0.8}{u^2}$

$u^2 = \dfrac{0.8}{0.2} = 4$

$u = 2 \text{ m/s}$

Thus, the minimum horizontal velocity required is $2 \text{ m/s}$.

Answer: $2$