A staircase has five steps each 10 cm high and 10 cm wide. What is the minimum horizontal velocity to be given to the ball, so that it hits directly the lowest plane from the top of the staircase? ig = 10 ms-2)

To hit the lowest plane (the 5th step) directly from the top, the ball must clear the edge of the 4th step. The staircase has 5 steps, each 10 cm (0.1 m) high and 10 cm (0.1 m) wide. To clear the 4th step, the ball must travel a horizontal distance (x) of at least 40 cm (0.4 m) while falling a vertical distance (y) of 40 cm (0.4 m). Using the kinematic equation for vertical motion, y = 0.5 * g * t², we find 0.4 = 0.5 * 10 * t², giving t = √0.08 seconds. For horizontal motion, x = u * t, so 0.4 = u * √0.08. Squaring both sides, 0.16 = u² * 0.08, which results in u² = 2, or u = √2 m/s. However, to ensure it lands on the 5th step and not the 4th, the velocity must be slightly greater than √2. Among the options, 2 m/s is the minimum integer velocity that ensures it clears the intermediate steps.