A sphere rolls off the top of a stairway with a horizontal velocity of magnitude 200 cm/sec. The steps are 10 cm high and 10 cm wide. Which step will the ball hit first?(g=10 `m//s^2`)

Options

A. 8
B. 2
C. 4
D. 6

Answer

To solve the problem of determining which step the sphere will hit first as it rolls off a stairway, we can break down the solution into several steps:

### Step 1: Understand the problem
The sphere rolls off the top of a stairway with a horizontal velocity of 200 cm/s. Each step is 10 cm high and 10 cm wide. We need to find out how many steps down the stairway the sphere will hit first.

### Step 2: Convert units
Since the height and width of the steps are given in centimeters, we should convert these measurements into meters for consistency with the gravitational acceleration given in m/s².

- Height of each step: \( h = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Width of each step: \( w = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Horizontal velocity: \( v = 200 \, \text{cm/s} = 2 \, \text{m/s} \)

### Step 3: Calculate the time taken to fall a certain height
The vertical motion of the sphere can be analyzed using the second equation of motion for free fall:

\[ s = \frac{1}{2} g t^2 \]

Where:
- \( s \) is the vertical distance fallen,
- \( g \) is the acceleration due to gravity (10 m/s²),
- \( t \) is the time in seconds.

For \( n \) steps, the height fallen is:

\[ s = n \times 0.1 \, \text{m} = 0.1n \, \text{m} \]

Substituting this into the equation gives:

\[ 0.1n = \frac{1}{2} \times 10 \times t^2 \]

This simplifies to:

\[ 0.1n = 5t^2 \quad \Rightarrow \quad t^2 = \frac{0.1n}{5} = \frac{n}{50} \]

### Step 4: Calculate the horizontal distance traveled
The horizontal distance \( x \) traveled by the sphere in time \( t \) is given by:

\[ x = v \times t = 2 \times t \]

Substituting \( t \) from the previous step:

\[ x = 2 \times \sqrt{\frac{n}{50}} = \frac{2\sqrt{n}}{\sqrt{50}} = \frac{2\sqrt{n}}{5\sqrt{2}} = \frac{2\sqrt{n}}{7.07} \approx 0.282\sqrt{n} \, \text{m} \]

### Step 5: Determine the relationship between horizontal distance and steps
The horizontal distance \( x \) must be less than or equal to the width of the steps multiplied by the number of steps \( n \):

\[ x \leq n \times 0.1 \, \text{m} \]

Substituting for \( x \):

\[ 0.282\sqrt{n} \leq 0.1n \]

### Step 6: Rearranging the inequality
Rearranging gives:

\[ 0.282\sqrt{n} \leq 0.1n \quad \Rightarrow \quad 0.282 \leq 0.1\sqrt{n} \quad \Rightarrow \quad \sqrt{n} \geq \frac{0.282}{0.1} = 2.82 \]

Squaring both sides:

\[ n \geq (2.82)^2 \approx 7.9524 \]

Since \( n \) must be a whole number, we round up to the nearest whole number:

\[ n \geq 8 \]

### Conclusion
The sphere will hit the 8th step first.