A particle is projected from the ground with an initial speed of v at an angle `theta` with horizontal. The average velocity of the particle between its point of projection and highest point of trajectroy is :

Options

A. `v/2sqrt(1+2 cos^(2) theta)`
B. `v/2sqrt(1+2 cos^(2) theta)`
C. `v/2sqrt(1+3 cos^(2) theta)`
D. `vcos theta`

Explanation

To find the average velocity of a particle projected from the ground with an initial speed \( v \) at an angle \( \theta \) with the horizontal, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The particle is projected at an angle \( \theta \) with an initial speed \( v \).
- The motion is a projectile motion, and we need to find the average velocity from the point of projection to the highest point of the trajectory.

2. **Formula for Average Velocity**:
- The average velocity \( \overline{v} \) is given by the formula:
\[ \overline{v} = \frac{\text{Total Displacement}}{\text{Total Time}} \]

3. **Finding Total Displacement**:
- The total displacement from the point of projection (A) to the highest point (B) is the vertical distance (height) at the highest point since the horizontal displacement does not contribute to the vertical component.
- The height \( H \) at the highest point can be calculated using the formula:
\[ H = \frac{v^2 \sin^2 \theta}{2g} \]
- The horizontal displacement at the highest point is \( R/2 \), where \( R \) is the total range. The range \( R \) can be calculated as:
\[ R = \frac{v^2 \sin 2\theta}{g} \]
- Thus, the horizontal displacement at the highest point is:
\[ r_1 = \frac{R}{2} = \frac{v^2 \sin 2\theta}{2g} \]

4. **Calculating Total Time**:
- The time taken to reach the highest point is half of the total time of flight. The total time of flight \( T \) can be calculated as:
\[ T = \frac{2v \sin \theta}{g} \]
- Therefore, the time to reach the highest point \( t \) is:
\[ t = \frac{T}{2} = \frac{v \sin \theta}{g} \]

5. **Substituting Values**:
- Now, substituting the values of total displacement and total time into the average velocity formula:
\[ \overline{v} = \frac{H}{t} \]
- Substitute \( H \) and \( t \):
\[ \overline{v} = \frac{\frac{v^2 \sin^2 \theta}{2g}}{\frac{v \sin \theta}{g}} = \frac{v \sin \theta}{2} \]

6. **Final Result**:
- Thus, the average velocity of the particle between its point of projection and the highest point of the trajectory is:
\[ \overline{v} = \frac{v \sin \theta}{2} \]