Just after firing, a bullet is found to move at an angle of `37^(@)` to the horizontal. Its acceleration is `10 m//s^(2)` downwards. Find the component of its acceleration in the direction of its velocity.

Options

A. `-6m//s^(2)`
B. `-4m//s^(2)`
C. `-8m//s^(2)`
D. `-5m//s^(2)`

Answer

To find the component of the bullet's acceleration in the direction of its velocity, we can follow these steps:

### Step 1: Understand the Problem
The bullet moves at an angle of \(37^\circ\) to the horizontal, and the acceleration due to gravity is \(10 \, \text{m/s}^2\) acting downwards. We need to find the component of this acceleration in the direction of the bullet's velocity.

### Step 2: Draw a Diagram
- Draw a horizontal line to represent the ground.
- Draw a line at an angle of \(37^\circ\) above the horizontal to represent the direction of the bullet's velocity.
- Draw a vertical line downwards to represent the acceleration due to gravity.

### Step 3: Identify Angles
- The angle between the acceleration vector (downward) and the velocity vector (at \(37^\circ\) to the horizontal) is \(53^\circ\) (since \(90^\circ - 37^\circ = 53^\circ\)).

### Step 4: Calculate the Component of Acceleration
To find the component of the acceleration vector in the direction of the velocity vector, we use the cosine of the angle between them. The formula for the component of acceleration \(a\) in the direction of velocity is:
\[ a_{\text{component}} = a \cdot \cos(\theta) \]
where:
- \(a = 10 \, \text{m/s}^2\) (the magnitude of acceleration),
- \(\theta = 53^\circ\) (the angle between the acceleration and the direction of velocity).

### Step 5: Substitute Values
Now, substituting the values into the formula:
\[ a_{\text{component}} = 10 \cdot \cos(53^\circ) \]

### Step 6: Calculate \(\cos(53^\circ)\)
Using the cosine value:
\[ \cos(53^\circ) = \frac{3}{5} \]
Thus,
\[ a_{\text{component}} = 10 \cdot \frac{3}{5} = 6 \, \text{m/s}^2 \]

### Step 7: Determine the Direction
Since the acceleration is acting downward and the velocity is directed upwards at \(37^\circ\), the component of acceleration in the direction of the velocity is negative:
\[ a_{\text{component}} = -6 \, \text{m/s}^2 \]

### Final Answer
The component of the bullet's acceleration in the direction of its velocity is:
\[ \boxed{-6 \, \text{m/s}^2} \]