The height and width of each step of a staircase are 20cm and 30cm respectively. A ball rolls off the top of a stair with horizontal velocity v and hits the fifth step. The magnitude of v is (g=10 `ms^(-2)`)

Options

A. 1.5`sqrt(5) ms^(-1)`
B. `3sqrt(5) ms^(-1)`
C. `7.5 ms^(-1)`
D. `1.5 ms^(-1)`

Answer

To solve the problem, we need to determine the horizontal velocity \( v \) of a ball that rolls off the top of a staircase and hits the fifth step. The height and width of each step are given as 20 cm and 30 cm, respectively. We will use the equations of motion to find the value of \( v \).

### Step-by-Step Solution:

1. **Identify the Total Vertical and Horizontal Distances**:
- The ball falls a total vertical distance \( y \) after hitting the fifth step.
- Since there are 5 steps, the total vertical distance is:
\[ y = 5 \times 20 \text{ cm} = 100 \text{ cm} = 1 \text{ m} \]
- The total horizontal distance \( x \) covered by the ball is:
\[ x = 5 \times 30 \text{ cm} = 150 \text{ cm} = 1.5 \text{ m} \]

2. **Use the Equation of Motion**:
- The equation relating vertical displacement \( y \), horizontal displacement \( x \), gravitational acceleration \( g \), and initial horizontal velocity \( v \) is given by:
\[ y = -\frac{g x^2}{2 v^2} \]
- Here, \( g = 10 \text{ m/s}^2 \).

3. **Substituting the Known Values**:
- Substitute \( y = 1 \text{ m} \), \( g = 10 \text{ m/s}^2 \), and \( x = 1.5 \text{ m} \) into the equation:
\[ 1 = -\frac{10 \times (1.5)^2}{2 v^2} \]

4. **Simplifying the Equation**:
- Calculate \( (1.5)^2 = 2.25 \):
\[ 1 = -\frac{10 \times 2.25}{2 v^2} \]
- This simplifies to:
\[ 1 = -\frac{22.5}{2 v^2} \]
- Rearranging gives:
\[ 2 v^2 = -22.5 \quad \Rightarrow \quad v^2 = \frac{22.5}{2} = 11.25 \]

5. **Finding the Value of \( v \)**:
- Taking the square root:
\[ v = \sqrt{11.25} = \sqrt{(5 \times 2.25)} = \sqrt{5} \times 1.5 \]
- Thus, we have:
\[ v = 1.5 \sqrt{5} \text{ m/s} \]

### Final Answer:
The magnitude of the horizontal velocity \( v \) is:
\[ v = 1.5 \sqrt{5} \text{ m/s} \]