A rifle is used to shoot twice at a target, using identical cartridges. The first time, the rifle is aimed parallel to the ground and directly at the center of the bull's-eye. The bullet strikes the target at a distance of `H_(A)` below the center, however. The second time, the rifle is similarly aimed, but from twice the distance from the target. This time the bullet strikes the target at a distance of `H_(B)` below the center. Find the ratio `H_(B)//H_(A)`.

Options

A. 2
B. 4
C. 3
D. 5

Answer

To solve the problem, we need to analyze the motion of the bullets in two different scenarios.

### Step 1: Understanding the Problem
We have two shots fired from a rifle aimed horizontally at a target. The first shot hits the target at a distance \( H_A \) below the center when fired from a distance \( D \). The second shot is fired from a distance \( 2D \) and hits the target at a distance \( H_B \) below the center. We need to find the ratio \( \frac{H_B}{H_A} \).

### Step 2: Analyzing the First Shot
For the first shot:
- The bullet is fired horizontally and falls under the influence of gravity.
- The time taken for the bullet to reach the target is \( T_A \).
- The vertical displacement due to gravity when the bullet reaches the target is given by:
\[ H_A = \frac{1}{2} g T_A^2 \]
- The horizontal distance covered by the bullet is:
\[ D = U_A T_A \]
where \( U_A \) is the initial horizontal velocity of the bullet.

### Step 3: Analyzing the Second Shot
For the second shot:
- The bullet is fired from a distance \( 2D \).
- The time taken for the bullet to reach the target is \( T_B \).
- The vertical displacement for this shot is:
\[ H_B = \frac{1}{2} g T_B^2 \]
- The horizontal distance covered is:
\[ 2D = U_A T_B \]

### Step 4: Relating the Distances and Times
From the equations for horizontal distance:
1. From the first shot:
\[ D = U_A T_A \quad \text{(1)} \]
2. From the second shot:
\[ 2D = U_A T_B \quad \text{(2)} \]

From equation (1), we can express \( T_A \):
\[ T_A = \frac{D}{U_A} \]

From equation (2), we can express \( T_B \):
\[ T_B = \frac{2D}{U_A} \]

### Step 5: Finding the Ratio of Times
Now, we can find the ratio of the times:
\[ \frac{T_B}{T_A} = \frac{\frac{2D}{U_A}}{\frac{D}{U_A}} = 2 \]

### Step 6: Finding the Ratio of \( H_B \) and \( H_A \)
Now substituting \( T_B \) and \( T_A \) into the equations for \( H_A \) and \( H_B \):
1. For \( H_A \):
\[ H_A = \frac{1}{2} g T_A^2 = \frac{1}{2} g \left(\frac{D}{U_A}\right)^2 \]
2. For \( H_B \):
\[ H_B = \frac{1}{2} g T_B^2 = \frac{1}{2} g \left(\frac{2D}{U_A}\right)^2 = \frac{1}{2} g \cdot 4 \left(\frac{D}{U_A}\right)^2 \]

### Step 7: Finding the Ratio \( \frac{H_B}{H_A} \)
Now we can find the ratio:
\[ \frac{H_B}{H_A} = \frac{\frac{1}{2} g \cdot 4 \left(\frac{D}{U_A}\right)^2}{\frac{1}{2} g \left(\frac{D}{U_A}\right)^2} = 4 \]

### Final Answer
Thus, the ratio \( \frac{H_B}{H_A} \) is:
\[ \frac{H_B}{H_A} = 4 \]