A bullet of mass m is fired at angle `theta` with the vertical. The bullet returns to the ground in time t. The total change of momentum equals

Options

A. `(mg)/t`
B. `(2mg)/t`
C. `m"gt"`
D. `2m"gt"`

Answer

To find the total change of momentum of a bullet fired at an angle \(\theta\) with the vertical and returning to the ground in time \(t\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The bullet is fired at an angle \(\theta\) with the vertical. This means it will follow a projectile motion path and will return to the ground after time \(t\).

2. **Identify Forces Acting on the Bullet**:
- The only force acting on the bullet during its flight is the gravitational force, which is given by \(F = mg\), where \(m\) is the mass of the bullet and \(g\) is the acceleration due to gravity.

3. **Change in Momentum**:
- According to Newton's second law, the change in momentum (\(\Delta p\)) can be calculated using the formula:
\[ \Delta p = F \cdot \Delta t \]
- Here, \(\Delta t\) is the total time of flight, which is given as \(t\).

4. **Substituting the Values**:
- Since the only force acting on the bullet is the gravitational force \(mg\), we can substitute this into the equation:
\[ \Delta p = mg \cdot t \]

5. **Final Result**:
- Thus, the total change of momentum of the bullet as it returns to the ground is:
\[ \Delta p = mgt \]

### Final Answer:
The total change of momentum equals \(mgt\).