A body is projected with an initial Velocity 20 m/s at 60° to the horizontal. Its initial velocity vector is- (g=10 `m//s^2`)

Options

A. `10hati -20 hatj`
B. `10sqrt(3)hati + 10 hatj`
C. `10hati + 10sqrt(3)hatj`
D. `5hati + 5sqrt(3)hatj`

Answer

To solve the problem of a body projected with an initial velocity of 20 m/s at an angle of 60° to the horizontal, we need to break down the initial velocity into its horizontal (x) and vertical (y) components.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial velocity (u) = 20 m/s
- Angle of projection (θ) = 60°
- Acceleration due to gravity (g) = 10 m/s² (though we won't need it for finding the initial velocity components)

2. **Calculate the Horizontal Component of Velocity (u_x):**
- The horizontal component can be calculated using the cosine function:
\[ u_x = u \cdot \cos(θ) \]
- Substitute the values:
\[ u_x = 20 \cdot \cos(60°) \]
- Since \(\cos(60°) = \frac{1}{2}\):
\[ u_x = 20 \cdot \frac{1}{2} = 10 \text{ m/s} \]

3. **Calculate the Vertical Component of Velocity (u_y):**
- The vertical component can be calculated using the sine function:
\[ u_y = u \cdot \sin(θ) \]
- Substitute the values:
\[ u_y = 20 \cdot \sin(60°) \]
- Since \(\sin(60°) = \frac{\sqrt{3}}{2}\):
\[ u_y = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \text{ m/s} \]

4. **Form the Initial Velocity Vector:**
- The initial velocity vector can be expressed in terms of its components:
\[ \vec{u} = u_x \hat{i} + u_y \hat{j} \]
- Substitute the values of \(u_x\) and \(u_y\):
\[ \vec{u} = 10 \hat{i} + 10\sqrt{3} \hat{j} \text{ m/s} \]

5. **Final Answer:**
- The initial velocity vector is:
\[ \vec{u} = 10 \hat{i} + 10\sqrt{3} \hat{j} \text{ m/s} \]