Two paper screens `A` and `B` are separated by a distance of `100m`. A bullet pierces `A` and `B`. The hole in `B` is `10 cm` below the hole in `A`. If the bullet is travelling horizontally at the time of hitting the screen `A`, calculate the velocity of the bullet when it hits the screen `A`. Neglect resistance of paper and air.

Options

A. `100 ms^(-1)`
B. `200 ms^(-1)`
C. `600 ms^(-1)`
D. `700 ms^(-1)`

Explanation

To solve the problem, we need to determine the velocity of the bullet when it hits screen A. We will use the principles of projectile motion, specifically focusing on the vertical motion of the bullet as it travels between the two screens.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Distance between screens A and B, \( d = 100 \, \text{m} \)
- Vertical drop from A to B, \( h = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Determine the Time of Fall:**
The bullet falls a vertical distance \( h \) while it travels horizontally. We can use the equation of motion for vertical displacement:
\[ h = \frac{1}{2} g t^2 \]
Rearranging this equation to solve for time \( t \):
\[ t^2 = \frac{2h}{g} \]
\[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 0.1}{10}} = \sqrt{0.02} = 0.1414 \, \text{s} \]

3. **Calculate the Horizontal Velocity:**
The bullet travels horizontally a distance of \( d \) in the same time \( t \):
\[ d = v_A \cdot t \]
Rearranging to find the horizontal velocity \( v_A \):
\[ v_A = \frac{d}{t} = \frac{100}{0.1414} \approx 707.1 \, \text{m/s} \]

4. **Final Result:**
The velocity of the bullet when it hits screen A is approximately \( 707.1 \, \text{m/s} \). For practical purposes, we can round this to \( 700 \, \text{m/s} \).

### Summary:
The velocity of the bullet when it hits screen A is approximately \( 700 \, \text{m/s} \).