A gun mounted on the top of a moving truck t is aimed in the backward direction at an angle of 30° to the vertical. If the muzzle velocity of the bullet is `4 ms^(-1)` the value of speed of the truck that will make the bullet come of out vertically is

Options

A. `1 ms^(-1)`
B. `sqrt(3)/2 ms^(-1)`
C. `0.5 ms^(-1)`
D. `2 ms^(-1)`

Answer

To solve the problem, we need to analyze the motion of the bullet fired from the gun mounted on the truck. The bullet is aimed backward at an angle of 30° to the vertical, and we need to find the speed of the truck such that the bullet exits vertically.

### Step-by-Step Solution:

1. **Understand the Problem**:
- The bullet is fired at an angle of 30° to the vertical.
- The muzzle velocity of the bullet is \( u = 4 \, \text{m/s} \).
- We need to determine the speed of the truck \( v_t \) such that the bullet exits vertically.

2. **Break Down the Bullet's Velocity**:
- The bullet's velocity can be resolved into two components:
- Vertical component: \( u_y = u \cos(30°) \)
- Horizontal component: \( u_x = u \sin(30°) \)

3. **Calculate the Components**:
- Using \( \cos(30°) = \frac{\sqrt{3}}{2} \) and \( \sin(30°) = \frac{1}{2} \):
- \( u_y = 4 \cos(30°) = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \, \text{m/s} \)
- \( u_x = 4 \sin(30°) = 4 \times \frac{1}{2} = 2 \, \text{m/s} \)

4. **Condition for Vertical Exit**:
- For the bullet to exit vertically, the horizontal component of the bullet's velocity must be equal to the speed of the truck moving in the opposite direction.
- Therefore, we set:
\[ v_t = u_x \]

5. **Substituting the Values**:
- From the previous calculation, we found \( u_x = 2 \, \text{m/s} \).
- Thus, the speed of the truck \( v_t \) is:
\[ v_t = 2 \, \text{m/s} \]

### Final Answer:
The speed of the truck that will make the bullet come out vertically is \( 2 \, \text{m/s} \).

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