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 and vertical components.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial velocity (u) = 60 m/s
- Angle of projection (θ) = 30 degrees

2. **Determine the Components of the Velocity:**
- The horizontal component (u_x) can be calculated using the formula:
\[ u_x = u \cdot \cos(θ) \]
- The vertical component (u_y) can be calculated using the formula:
\[ u_y = u \cdot \sin(θ) \]

3. **Calculate the Horizontal Component:**
- Using the cosine of 30 degrees:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]
- Therefore,
\[ u_x = 60 \cdot \cos(30^\circ) = 60 \cdot \frac{\sqrt{3}}{2} = 30\sqrt{3} \text{ m/s} \]

4. **Calculate the Vertical Component:**
- Using the sine of 30 degrees:
\[ \sin(30^\circ) = \frac{1}{2} \]
- Therefore,
\[ u_y = 60 \cdot \sin(30^\circ) = 60 \cdot \frac{1}{2} = 30 \text{ m/s} \]

5. **Write the Velocity Vector:**
- The initial velocity vector can be expressed in terms of its components as:
\[ \vec{u} = u_x \hat{i} + u_y \hat{j} \]
- Substituting the values we calculated:
\[ \vec{u} = 30\sqrt{3} \hat{i} + 30 \hat{j} \]

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