A body is projected with a velocity `60 ms ^(-1) at 30^(@)` to horizontal . Its initial velocity vector is

Options

A. `10 hati + 10 sqrt3 hatj `
B. ` 30 hati + 30 sqrt3 hatj`
C. `30sqrt(3hati) + 30 hatj`
D. `30sqrt(3hati)`

Answer

To find the initial velocity vector of a body projected at a velocity of 60 m/s at an angle of 30 degrees to the horizontal, we can break down the velocity into its horizontal (x) and vertical (y) components using trigonometric functions.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial velocity \( V = 60 \, \text{m/s} \)
- Angle of projection \( \theta = 30^\circ \)

2. **Calculate the Horizontal Component (Vx):**
- The horizontal component of the velocity can be calculated using the cosine function:
\[ V_x = V \cdot \cos(\theta) = 60 \cdot \cos(30^\circ) \]
- The value of \( \cos(30^\circ) \) is \( \frac{\sqrt{3}}{2} \).
\[ V_x = 60 \cdot \frac{\sqrt{3}}{2} = 30\sqrt{3} \, \text{m/s} \]

3. **Calculate the Vertical Component (Vy):**
- The vertical component of the velocity can be calculated using the sine function:
\[ V_y = V \cdot \sin(\theta) = 60 \cdot \sin(30^\circ) \]
- The value of \( \sin(30^\circ) \) is \( \frac{1}{2} \).
\[ V_y = 60 \cdot \frac{1}{2} = 30 \, \text{m/s} \]

4. **Combine the Components to Form the Velocity Vector:**
- The initial velocity vector \( \vec{V} \) can be expressed in terms of its components:
\[ \vec{V} = V_x \hat{i} + V_y \hat{j} \]
- Substituting the values we calculated:
\[ \vec{V} = 30\sqrt{3} \hat{i} + 30 \hat{j} \]

5. **Final Answer:**
- The initial velocity vector is:
\[ \vec{V} = 30\sqrt{3} \hat{i} + 30 \hat{j} \, \text{m/s} \]