Question

A ball rolls off the top of a stairway with a horizontal velocity $u = \sqrt{x}$ m/s. The steps are 0.2 m high and 0.2 m wide. The ball just hits the edge of the 4th step. The value of $x$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

3

Answer 2

The ball follows projectile motion. To hit the edge of the 4th step, it must cover a horizontal distance of $(4-1) \times 0.2 = 0.6$ m and fall a vertical distance of $(4-1) \times 0.2 = 0.6$ m.

Using the vertical motion equation: $H = \frac{1}{2}gt^2$ $0.6 = \frac{1}{2} \times 10 \times t^2$ $t^2 = \frac{0.6}{5} = \frac{3}{25}$ $t = \frac{\sqrt{3}}{5}$ s.

Using the horizontal motion equation: $R = ut$ $0.6 = u \times \frac{\sqrt{3}}{5}$ $u = \frac{0.6 \times 5}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$ m/s.

Given $u = \sqrt{x}$, we have $\sqrt{x} = \sqrt{3}$, so $x = 3$.