A particle is projected at an angle of `30^(@)w.r.t.` horizontal with speed `20m//s:(` use `g=10m//s^(2))`
`(i)` Find the position vector of the particle after `1s`.
`(ii)` Find the angle between velocity vector and position vector at `t=1s`.

Explanation

To solve the problem step by step, we will break it down into two parts as given in the question.

### Part (i): Find the position vector of the particle after 1s.

1. **Identify the components of the initial velocity:**
- The particle is projected at an angle of \(30^\circ\) with an initial speed of \(20 \, \text{m/s}\).
- The horizontal component of the velocity (\(V_x\)) is given by:
\[ V_x = V \cos(\theta) = 20 \cos(30^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]
- The vertical component of the velocity (\(V_y\)) is given by:
\[ V_y = V \sin(\theta) = 20 \sin(30^\circ) = 20 \times \frac{1}{2} = 10 \, \text{m/s} \]

2. **Calculate the position after 1 second:**
- The horizontal position after \(t = 1 \, \text{s}\) is:
\[ r_x = V_x \cdot t = 10\sqrt{3} \cdot 1 = 10\sqrt{3} \, \text{m} \]
- The vertical position after \(t = 1 \, \text{s}\) is calculated using the equation of motion:
\[ r_y = V_y \cdot t - \frac{1}{2} g t^2 \]
where \(g = 10 \, \text{m/s}^2\):
\[ r_y = 10 \cdot 1 - \frac{1}{2} \cdot 10 \cdot (1)^2 = 10 - 5 = 5 \, \text{m} \]

3. **Combine the components to form the position vector:**
- The position vector \( \mathbf{r} \) after 1 second is:
\[ \mathbf{r} = r_x \hat{i} + r_y \hat{j} = 10\sqrt{3} \hat{i} + 5 \hat{j} \, \text{m} \]

### Part (ii): Find the angle between the velocity vector and position vector at \(t = 1s\).

1. **Determine the velocity vector at \(t = 1 \, \text{s}\):**
- The horizontal component remains constant:
\[ V_x = 10\sqrt{3} \, \text{m/s} \]
- The vertical component after \(t = 1 \, \text{s}\) is:
\[ V_y = V_y - gt = 10 - 10 \cdot 1 = 0 \, \text{m/s} \]
- Therefore, the velocity vector \( \mathbf{V} \) is:
\[ \mathbf{V} = 10\sqrt{3} \hat{i} + 0 \hat{j} = 10\sqrt{3} \hat{i} \, \text{m/s} \]

2. **Calculate the angle between the position vector and the velocity vector using the dot product:**
- The dot product \( \mathbf{V} \cdot \mathbf{r} \) is:
\[ \mathbf{V} \cdot \mathbf{r} = (10\sqrt{3} \hat{i}) \cdot (10\sqrt{3} \hat{i} + 5 \hat{j}) = 10\sqrt{3} \cdot 10\sqrt{3} + 0 = 300 \]
- The magnitudes of the vectors are:
\[ |\mathbf{V}| = 10\sqrt{3} \quad \text{and} \quad |\mathbf{r}| = \sqrt{(10\sqrt{3})^2 + 5^2} = \sqrt{300 + 25} = \sqrt{325} \]

3. **Using the dot product formula:**
\[ \mathbf{V} \cdot \mathbf{r} = |\mathbf{V}| |\mathbf{r}| \cos(\theta) \]
Substituting the values:
\[ 300 = (10\sqrt{3}) \cdot \sqrt{325} \cos(\theta) \]
Rearranging gives:
\[ \cos(\theta) = \frac{300}{10\sqrt{3} \cdot \sqrt{325}} = \frac{30}{\sqrt{3} \cdot \sqrt{325}} = \frac{10\sqrt{3}}{\sqrt{325}} \]

4. **Finding the angle:**
\[ \theta = \cos^{-1}\left(\frac{10\sqrt{3}}{\sqrt{325}}\right) \]

### Final Answers:
(i) The position vector after 1 second is \( \mathbf{r} = 10\sqrt{3} \hat{i} + 5 \hat{j} \, \text{m} \).

(ii) The angle between the velocity vector and position vector at \(t = 1s\) is \( \theta = \cos^{-1}\left(\frac{10\sqrt{3}}{\sqrt{325}}\right) \).