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

Correct Answer: 2

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: \(x = 2\)