A particle is projected with an initial velocity of `200 m//s` in a direction making an angle of `30^(@)` with the vertical. The horizontal distance covered by the particle in `3s` is

Options

A. `300 m`
B. `150 m`
C. `175 m`
D. `125 m`

Explanation

To solve the problem of finding the horizontal distance covered by a particle projected with an initial velocity of 200 m/s at an angle of 30 degrees with the vertical, we can follow these steps:

### Step 1: Understand the components of the initial velocity
The initial velocity \( v_0 = 200 \, \text{m/s} \) is given at an angle of \( 30^\circ \) with the vertical. To find the horizontal and vertical components of the velocity, we can use trigonometric functions.

- The angle with the horizontal will be \( 90^\circ - 30^\circ = 60^\circ \).

### Step 2: Calculate the horizontal component of the velocity
The horizontal component \( v_{x} \) can be calculated using:
\[ v_{x} = v_0 \cdot \cos(60^\circ) \]
Substituting the values:
\[ v_{x} = 200 \cdot \cos(60^\circ) = 200 \cdot \frac{1}{2} = 100 \, \text{m/s} \]

### Step 3: Use the horizontal component to find the distance
The horizontal distance \( x \) covered in time \( t \) can be calculated using the formula:
\[ x = v_{x} \cdot t \]
Given \( t = 3 \, \text{s} \):
\[ x = 100 \cdot 3 = 300 \, \text{m} \]

### Conclusion
The horizontal distance covered by the particle in 3 seconds is \( 300 \, \text{meters} \).

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