Two paper screens A and B are separated by distance 100 m. A bullet penetrates A and B, at points P and Q respectively, where Q is 10 cm below P. If bullet is travelling horizotally at the time of hitting A, the velocity of bullet at A is nearly

Options

A. 100 m/s
B. 200 m/s
C. 600 m/s
D. 700 m/s

Answer

To solve the problem, we need to analyze the motion of the bullet as it travels horizontally and falls vertically due to gravity. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have two screens A and B separated by a distance of 100 m. The bullet travels horizontally and penetrates both screens at points P and Q, where Q is 10 cm below P. We need to find the velocity of the bullet at screen A.

### Step 2: Identify the Vertical Motion
The bullet falls vertically due to gravity while traveling horizontally. The vertical distance fallen (from P to Q) is 10 cm, which is equal to 0.1 m.

### Step 3: Use the Equation of Motion
We can use the second equation of motion for vertical displacement:
\[ s = ut + \frac{1}{2} a t^2 \]
Here,
- \(s\) is the vertical displacement (0.1 m),
- \(u\) is the initial vertical velocity (0 m/s, since it starts falling from rest),
- \(a\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)),
- \(t\) is the time taken to fall.

Substituting the values, we have:
\[ 0.1 = 0 + \frac{1}{2} \cdot 9.81 \cdot t^2 \]
This simplifies to:
\[ 0.1 = 4.905 t^2 \]

### Step 4: Solve for Time \(t\)
Rearranging the equation gives:
\[ t^2 = \frac{0.1}{4.905} \]
Calculating this:
\[ t^2 \approx 0.0204 \quad \Rightarrow \quad t \approx \sqrt{0.0204} \approx 0.142 \, \text{s} \]

### Step 5: Calculate Horizontal Velocity
Now, we know the bullet travels horizontally a distance of 100 m in the time \(t\). Using the formula:
\[ \text{Distance} = \text{Velocity} \times \text{Time} \]
We can rearrange it to find the horizontal velocity \(v\):
\[ v = \frac{\text{Distance}}{t} = \frac{100}{0.142} \]
Calculating this gives:
\[ v \approx 704.2 \, \text{m/s} \]

### Step 6: Final Answer
Thus, the velocity of the bullet at screen A is approximately **700 m/s** (rounded to the nearest hundred).