Two paper screens A and B are separated by 150m. A bullet pierces A and B. The hole in B in 15 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: `(g=10ms^(-2))`

Options

A. `100 sqrt(3) m s^-1`
B. `200 sqrt(3) m s^-1`
C. `300 sqrt(3) m s^-1`
D. `500 sqrt(3) m s^-1`

Explanation

To solve the problem step by step, we will analyze the motion of the bullet as it travels horizontally and vertically between the two screens A and B.

### Step 1: Understand the scenario
The bullet travels horizontally and pierces through two screens A and B, which are separated by a vertical distance of 15 cm (or 0.15 m) and a horizontal distance of 150 m.

### Step 2: Calculate the time of flight
Since the bullet is traveling horizontally, we can use the equations of motion to find the time it takes for the bullet to fall 0.15 m vertically due to gravity. The vertical motion can be described by the equation:

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

Where:
- \( h = 0.15 \, \text{m} \) (the vertical distance fallen)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t \) is the time of flight.

Rearranging the equation to solve for \( t \):

\[ t^2 = \frac{2h}{g} \]

Substituting the values:

\[ t^2 = \frac{2 \times 0.15}{10} = \frac{0.3}{10} = 0.03 \]

Taking the square root:

\[ t = \sqrt{0.03} = \sqrt{3 \times 10^{-2}} = \sqrt{3}/10 \, \text{s} \]

### Step 3: Calculate the horizontal velocity
The horizontal distance traveled by the bullet is 150 m. The horizontal velocity \( u \) can be calculated using the formula:

\[ u = \frac{\text{distance}}{\text{time}} = \frac{150}{t} \]

Substituting \( t \):

\[ u = \frac{150}{\sqrt{3}/10} = 150 \times \frac{10}{\sqrt{3}} = \frac{1500}{\sqrt{3}} \, \text{m/s} \]

### Step 4: Simplify the expression
To express the velocity in a more standard form, we can multiply the numerator and the denominator by \( \sqrt{3} \):

\[ u = \frac{1500 \sqrt{3}}{3} = 500 \sqrt{3} \, \text{m/s} \]

### Final Answer
Thus, the velocity of the bullet at screen A is:

\[ u = 500 \sqrt{3} \, \text{m/s} \]