A body is projected with velocity 60m/sec at 30 degree horizontal, its initial velocity vector is ? In the above problem velocity after 3 seconds is-

Options

A. `20hati + 20sqrt(3) hatj`
B. `30hati`
C. `10sqrt(3)hatj`
D. `30sqrt(3)hati`

Answer

To solve the problem step by step, we will break it down into two parts: finding the initial velocity vector and then calculating the velocity after 3 seconds.

### Step 1: Determine the Initial Velocity Vector

1. **Identify the components of the initial velocity**:
- The body is projected with a velocity \( u = 60 \, \text{m/s} \) at an angle \( \theta = 30^\circ \) with respect to the horizontal.

2. **Calculate the horizontal and vertical components of the initial velocity**:
- The horizontal component \( u_x \) is given by:
\[ u_x = u \cos \theta = 60 \cos 30^\circ \]
- The vertical component \( u_y \) is given by:
\[ u_y = u \sin \theta = 60 \sin 30^\circ \]

3. **Substituting the values of \( \cos 30^\circ \) and \( \sin 30^\circ \)**:
- We know that \( \cos 30^\circ = \frac{\sqrt{3}}{2} \) and \( \sin 30^\circ = \frac{1}{2} \).
- Thus,
\[ u_x = 60 \times \frac{\sqrt{3}}{2} = 30\sqrt{3} \, \text{m/s} \]
\[ u_y = 60 \times \frac{1}{2} = 30 \, \text{m/s} \]

4. **Write the initial velocity vector**:
- The initial velocity vector \( \mathbf{u} \) can be expressed as:
\[ \mathbf{u} = u_x \hat{i} + u_y \hat{j} = 30\sqrt{3} \hat{i} + 30 \hat{j} \]

### Step 2: Calculate the Velocity After 3 Seconds

1. **Define the acceleration**:
- The only acceleration acting on the body is due to gravity, which acts downward. Therefore, the acceleration vector \( \mathbf{a} \) is:
\[ \mathbf{a} = -g \hat{j} = -10 \hat{j} \, \text{m/s}^2 \]

2. **Use the equation of motion to find the velocity after 3 seconds**:
- The velocity \( \mathbf{v} \) after time \( t \) is given by:
\[ \mathbf{v} = \mathbf{u} + \mathbf{a} t \]
- Substituting the known values:
\[ \mathbf{v} = (30\sqrt{3} \hat{i} + 30 \hat{j}) + (-10 \hat{j}) \cdot 3 \]
- This simplifies to:
\[ \mathbf{v} = 30\sqrt{3} \hat{i} + 30 \hat{j} - 30 \hat{j} \]
- Thus, we have:
\[ \mathbf{v} = 30\sqrt{3} \hat{i} + 0 \hat{j} \]

3. **Final result for the velocity after 3 seconds**:
- Therefore, the velocity after 3 seconds is:
\[ \mathbf{v} = 30\sqrt{3} \hat{i} \]

### Summary of Results
- The initial velocity vector is \( \mathbf{u} = 30\sqrt{3} \hat{i} + 30 \hat{j} \).
- The velocity after 3 seconds is \( \mathbf{v} = 30\sqrt{3} \hat{i} \).