A particle is projected at an angle `theta =30^@` with the horizontal, with a velocity of `10ms^(-1)`. Then

Options

A. After 2 s, the velocity of particle makes an angle of `60^@` with initial velocity vector.
B. After 1 s, the velocity of particles makes an angle of `60^@` with initial velocity vector.
C. The magnitude of velocity of particle after 1 s is `10 ms^(-1)`.
D. The magnitude of velocity of particle after 1 s is `5 ms^(-1)`.

Explanation

To solve the problem step by step, we will analyze the motion of the particle projected at an angle of \( \theta = 30^\circ \) with an initial velocity of \( 10 \, \text{m/s} \).

### Step 1: Determine the Initial Velocity Components
The initial velocity \( \vec{v_0} \) can be broken down into its horizontal and vertical components using trigonometric functions:
- \( v_{0x} = v_0 \cos(\theta) \)
- \( v_{0y} = v_0 \sin(\theta) \)

Given:
- \( v_0 = 10 \, \text{m/s} \)
- \( \theta = 30^\circ \)

Calculating the components:
- \( v_{0x} = 10 \cos(30^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \)
- \( v_{0y} = 10 \sin(30^\circ) = 10 \times \frac{1}{2} = 5 \, \text{m/s} \)

Thus, the initial velocity vector is:
\[ \vec{v_0} = 5\sqrt{3} \hat{i} + 5 \hat{j} \]

### Step 2: Determine the Velocity After 1 Second
The horizontal component of the velocity remains constant throughout the motion:
\[ v_x = v_{0x} = 5\sqrt{3} \, \text{m/s} \]

The vertical component of the velocity changes due to gravity:
\[ v_y = v_{0y} - g t \]
Where \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity) and \( t = 1 \, \text{s} \):
\[ v_y = 5 - 10 \times 1 = 5 - 10 = -5 \, \text{m/s} \]

Thus, the velocity vector after 1 second is:
\[ \vec{v} = 5\sqrt{3} \hat{i} - 5 \hat{j} \]

### Step 3: Calculate the Magnitude of the Velocity After 1 Second
The magnitude of the velocity vector is given by:
\[ |\vec{v}| = \sqrt{v_x^2 + v_y^2} \]
Substituting the values:
\[ |\vec{v}| = \sqrt{(5\sqrt{3})^2 + (-5)^2} = \sqrt{75 + 25} = \sqrt{100} = 10 \, \text{m/s} \]

### Step 4: Determine the Angle Between Initial and Final Velocity Vectors
To find the angle \( \theta \) between the initial and final velocity vectors, we use the dot product:
\[ \cos(\theta) = \frac{\vec{v_0} \cdot \vec{v}}{|\vec{v_0}| |\vec{v}|} \]

Calculating the dot product:
\[ \vec{v_0} \cdot \vec{v} = (5\sqrt{3})(5\sqrt{3}) + (5)(-5) = 75 - 25 = 50 \]

The magnitudes are:
- \( |\vec{v_0}| = 10 \, \text{m/s} \)
- \( |\vec{v}| = 10 \, \text{m/s} \)

Thus,
\[ \cos(\theta) = \frac{50}{10 \times 10} = \frac{50}{100} = 0.5 \]
Therefore,
\[ \theta = \cos^{-1}(0.5) = 60^\circ \]

### Conclusion
After 1 second, the velocity of the particle makes an angle of \( 60^\circ \) with the initial velocity vector, and the magnitude of the velocity after 1 second is \( 10 \, \text{m/s} \).