A particle is projected from a horizontal floor with speed `10 m//s` at an angle `30^(@)` with the floor and striking the floor after sometime. State which is correct.

Options

A. velocity of particle will be perpendicular to initial direction two seconds after projection.
B. minimum speed of particle will be `5 m//sec`.
C. Displacement of particle after half second will be `35//4m`.
D. None of these

Explanation

To solve the problem, we need to analyze the projectile motion of the particle projected at a speed of 10 m/s at an angle of 30 degrees with the horizontal. We will evaluate the statements given in the question and determine which one is correct.

### Step-by-Step Solution:

1. **Identify the Initial Velocity Components:**
- The initial velocity \( u = 10 \, \text{m/s} \).
- The angle of projection \( \theta = 30^\circ \).
- The horizontal component of the initial velocity \( u_x = u \cos \theta = 10 \cos 30^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \).
- The vertical component of the initial velocity \( u_y = u \sin \theta = 10 \sin 30^\circ = 10 \times \frac{1}{2} = 5 \, \text{m/s} \).

2. **Calculate the Time of Flight:**
- The time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{2u_y}{g} \]
- Here, \( g \) (acceleration due to gravity) is approximately \( 10 \, \text{m/s}^2 \).
- Substituting the values:
\[ T = \frac{2 \times 5}{10} = 1 \, \text{s} \]

3. **Evaluate the Statements:**
- **Option A:** "Velocity of particle will be perpendicular to the initial direction 2 seconds after projection."
- The time of flight is 1 second, so this statement is incorrect because the particle will not be in motion for 2 seconds.

- **Option B:** "Minimum speed of particle will be 5 m/s."
- The minimum speed occurs at the peak of the trajectory where the vertical component of velocity is zero. The horizontal component remains constant at \( 5\sqrt{3} \, \text{m/s} \), which is approximately \( 8.66 \, \text{m/s} \). Therefore, this statement is also incorrect.

- **Option C:** "Displacement of particle after half a second will be \( \frac{35}{4} \, \text{m} \)."
- To find the displacement after \( t = 0.5 \, \text{s} \):
- Horizontal displacement \( S_x = u_x \cdot t = 5\sqrt{3} \cdot 0.5 = \frac{5\sqrt{3}}{2} \).
- Vertical displacement \( S_y = u_y \cdot t - \frac{1}{2} g t^2 = 5 \cdot 0.5 - \frac{1}{2} \cdot 10 \cdot (0.5)^2 = 2.5 - 1.25 = 1.25 \, \text{m} \).
- The resultant displacement \( S = \sqrt{S_x^2 + S_y^2} \).
- Calculate \( S_x^2 = \left(\frac{5\sqrt{3}}{2}\right)^2 = \frac{75}{4} \) and \( S_y^2 = (1.25)^2 = \frac{25}{16} \).
- Thus, \( S = \sqrt{\frac{75}{4} + \frac{25}{16}} \).
- Finding a common denominator (16):
\[ S = \sqrt{\frac{300}{16} + \frac{25}{16}} = \sqrt{\frac{325}{16}} = \frac{\sqrt{325}}{4} \approx 4.03 \, \text{m} \]
- Since \( \frac{35}{4} = 8.75 \, \text{m} \), this statement is also incorrect.

4. **Conclusion:**
- Since all the provided options are incorrect, the correct answer is that none of these options are correct.

### Final Answer:
The correct option is: None of these.