Two paper screens A and B are separated by a distance of 200 m. A bullet pierces A and then B. The hole in B is 40 cm below the hole in A. If the bullet is travelling horizontally at the time of hitting A, then the velocity of the bullet at A is 

Options

A. `200 ms^(-1)`
B. `400 ms^(-1)`
C. `600 ms^(-1)`
D. `700 ms^(-1)`

Explanation

To solve the problem, we need to determine the horizontal velocity of the bullet at the moment it hits screen A. We can break down the solution into several steps:

### Step 1: Understand the Problem
We have two screens, A and B, separated by a distance of 200 m. The bullet travels horizontally and makes a hole in both screens. The hole in screen B is 40 cm (or 0.4 m) below the hole in screen A.

### Step 2: Determine the Time of Flight
Since the bullet is traveling horizontally, we can use the horizontal distance to calculate the time it takes to travel from screen A to screen B. The distance between the screens is 200 m.

Using the formula:
\[ \text{Distance} = \text{Velocity} \times \text{Time} \]
Let \( u \) be the horizontal velocity of the bullet. The time \( t \) taken to travel from A to B can be expressed as:
\[ t = \frac{200}{u} \]

### Step 3: Analyze the Vertical Motion
While the bullet is traveling horizontally, it is also falling due to gravity. The vertical distance fallen is 0.4 m. We can use the following kinematic equation for vertical motion:
\[ s = ut + \frac{1}{2} a t^2 \]
In this case, the initial vertical velocity \( u \) is 0 (since the bullet starts falling from rest), and the acceleration \( a \) is \( g = 9.8 \, \text{m/s}^2 \). Thus, the equation simplifies to:
\[ 0.4 = 0 + \frac{1}{2} \cdot 9.8 \cdot t^2 \]

### Step 4: Substitute for Time
Now we substitute \( t = \frac{200}{u} \) into the vertical motion equation:
\[ 0.4 = \frac{1}{2} \cdot 9.8 \cdot \left(\frac{200}{u}\right)^2 \]

### Step 5: Solve for \( u \)
Rearranging the equation gives:
\[ 0.4 = \frac{1}{2} \cdot 9.8 \cdot \frac{40000}{u^2} \]
Multiplying both sides by \( 2u^2 \) gives:
\[ 0.8u^2 = 9.8 \cdot 40000 \]
\[ u^2 = \frac{9.8 \cdot 40000}{0.8} \]
Calculating the right side:
\[ u^2 = \frac{392000}{0.8} = 490000 \]
Taking the square root:
\[ u = \sqrt{490000} \approx 700 \, \text{m/s} \]

### Final Answer
The velocity of the bullet at screen A is approximately \( 700 \, \text{m/s} \).

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