Two thin wood screens A and B are separated by 200m. A bullet travelling horizontally at a speed of 600`ms^(-1)` hits the screen A, penetrates through it and finally emerges out from B making holes in A and B. If the resistance of air and wood arc negligible, the difference of heights of the holes in A and B is

Options

A. 5 m
B. `49/90` m
C. `7/sqrt(90)` m
D. zero

Answer

To solve the problem step by step, we need to analyze the motion of the bullet as it travels horizontally and the effect of gravity on its vertical position.

### Step 1: Identify the given data
- Distance between screens A and B, \( D = 200 \, \text{m} \)
- Speed of the bullet, \( v = 600 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Calculate the time taken for the bullet to travel from A to B
Using the formula for time:
\[ t = \frac{D}{v} \]
Substituting the values:
\[ t = \frac{200 \, \text{m}}{600 \, \text{m/s}} = \frac{1}{3} \, \text{s} \]

### Step 3: Analyze the vertical motion of the bullet
Since the bullet is moving horizontally, its initial vertical velocity \( u = 0 \). The vertical displacement \( h \) after time \( t \) can be calculated using the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( a = g \) (acceleration due to gravity), so:
\[ h = 0 \cdot t + \frac{1}{2} g t^2 \]
Substituting \( t = \frac{1}{3} \, \text{s} \):
\[ h = \frac{1}{2} \cdot 9.8 \cdot \left(\frac{1}{3}\right)^2 \]
Calculating \( \left(\frac{1}{3}\right)^2 = \frac{1}{9} \):
\[ h = \frac{1}{2} \cdot 9.8 \cdot \frac{1}{9} = \frac{9.8}{18} = \frac{49}{90} \, \text{m} \]

### Step 4: Conclusion
The difference in heights of the holes in screens A and B is:
\[ h = \frac{49}{90} \, \text{m} \]