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 of a particle projected at an angle of \( \theta = 30^\circ \) with an initial velocity of \( 10 \, \text{m/s} \), we will calculate the velocity of the particle after 1 second and 2 seconds, as well as the angle of the velocity vector at those times.

### Step 1: Resolve the initial velocity into components
The initial velocity \( u \) can be resolved into horizontal and vertical components:
- \( u_x = u \cos \theta = 10 \cos 30^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \)
- \( u_y = u \sin \theta = 10 \sin 30^\circ = 10 \times \frac{1}{2} = 5 \, \text{m/s} \)

### Step 2: Calculate the vertical velocity after 1 second
The vertical component of the velocity after time \( t \) is given by:
\[ v_y = u_y - g t \]
where \( g \) is the acceleration due to gravity (\( g \approx 9.81 \, \text{m/s}^2 \)).
For \( t = 1 \, \text{s} \):
\[ v_y = 5 - 9.81 \times 1 = 5 - 9.81 = -4.81 \, \text{m/s} \]

### Step 3: Calculate the vertical velocity after 2 seconds
For \( t = 2 \, \text{s} \):
\[ v_y = 5 - 9.81 \times 2 = 5 - 19.62 = -14.62 \, \text{m/s} \]

### Step 4: Calculate the total velocity after 1 second
The total velocity \( v \) after 1 second can be calculated using the Pythagorean theorem:
\[ v = \sqrt{u_x^2 + v_y^2} \]
\[ v = \sqrt{(5\sqrt{3})^2 + (-4.81)^2} \]
\[ v = \sqrt{75 + 23.1361} = \sqrt{98.1361} \approx 9.9 \, \text{m/s} \]

### Step 5: Calculate the angle of the velocity vector after 1 second
The angle \( \phi \) of the velocity vector with respect to the horizontal can be found using:
\[ \tan \phi = \frac{v_y}{u_x} \]
\[ \tan \phi = \frac{-4.81}{5\sqrt{3}} \]
Calculating \( \phi \):
\[ \phi = \tan^{-1}\left(\frac{-4.81}{5\sqrt{3}}\right) \]
This will give an angle in the fourth quadrant (since \( v_y \) is negative).

### Step 6: Calculate the total velocity after 2 seconds
Using the same method as in Step 4:
\[ v = \sqrt{(5\sqrt{3})^2 + (-14.62)^2} \]
\[ v = \sqrt{75 + 214.4644} = \sqrt{289.4644} \approx 17.0 \, \text{m/s} \]

### Step 7: Calculate the angle of the velocity vector after 2 seconds
Using the same method as in Step 5:
\[ \tan \phi = \frac{-14.62}{5\sqrt{3}} \]
Calculating \( \phi \):
\[ \phi = \tan^{-1}\left(\frac{-14.62}{5\sqrt{3}}\right) \]

### Summary of Results
- After 1 second:
- Velocity: \( \approx 9.9 \, \text{m/s} \)
- Angle: \( \phi \) (calculated based on the above steps)

- After 2 seconds:
- Velocity: \( \approx 17.0 \, \text{m/s} \)
- Angle: \( \phi \) (calculated based on the above steps)