Question

A ball rolls off the top of a stairway with horizontal velocity $u$. The steps are $0.1 \, \text{m}$ high and $0.1 \, \text{m}$ wide. The minimum velocity $u$ with which the ball just hits the step 5 of the stairway will be $\sqrt{x} \, \text{m/s}$ where $x =$ _____. (Use $g = 10 \, \text{m/s}^2$).

Answer 1

Step 1: Determine Horizontal Range to Just Hit Step 5:

- The ball needs to cross 4 steps horizontally to just hit the 5th step. Since each step is 0.1 m wide, the horizontal range $R$ required to reach the 5th step is:

$R = 0.4 \text{ m}$

- Using the horizontal motion equation $R = u \cdot t$, we get:

$t = \frac{R}{u} = \frac{0.4}{u}$

Step 2: Vertical Motion Analysis:

- For vertical displacement, the ball needs to fall a height of $h = 4 \times 0.1 = 0.4 \text{ m}$. Using the vertical motion equation $h = \frac{1}{2} g t^2$:

$0.4 = \frac{1}{2} \cdot 10 \cdot \left( \frac{0.4}{u} \right)^2$

- Simplify to find $u$:

$0.4 = 5 \cdot \frac{0.16}{u^2}$

$u^2 = 2$

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

Step 3: Determine $x$:

- Given that $u = \sqrt{x}$, we find $x = 2$.

So, the correct answer is: $x = 2$