Average velocity of a particle in projectile motion between its starting point and the highest point of its trajectory is (projectin speed = u, angle projection from horizontal =`theta`)

Options

A. `u cos theta`
B. `u/2 sqrt(1+ cos^2 theta)`
C. `u/2 sqrt(2+ cos^2 theta)`
D. `u/2 sqrt(1+cos^2 t heta)`

Explanation

To find the average velocity of a particle in projectile motion between its starting point and the highest point of its trajectory, we can follow these steps:

### Step 1: Understand the Projectile Motion
When a particle is projected at an angle \( \theta \) with an initial speed \( u \), it follows a parabolic trajectory. The highest point of the trajectory is where the vertical component of the velocity becomes zero.

### Step 2: Determine the Displacement
1. **Horizontal Displacement**: At the highest point, the horizontal displacement is half of the total range. The total range \( R \) for projectile motion is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Therefore, the horizontal displacement to the highest point is:
\[ \text{Horizontal Displacement} = \frac{R}{2} = \frac{u^2 \sin 2\theta}{2g} \]

2. **Vertical Displacement**: The vertical displacement at the highest point is the maximum height \( H \), which can be calculated as:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

### Step 3: Calculate the Net Displacement
The net displacement \( S \) from the starting point to the highest point can be found using the Pythagorean theorem:
\[ S = \sqrt{(\text{Horizontal Displacement})^2 + (\text{Vertical Displacement})^2} \]
Substituting the values:
\[ S = \sqrt{\left(\frac{u^2 \sin 2\theta}{2g}\right)^2 + \left(\frac{u^2 \sin^2 \theta}{2g}\right)^2} \]

### Step 4: Simplify the Expression
1. Substitute \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ S = \sqrt{\left(\frac{u^2 (2 \sin \theta \cos \theta)}{2g}\right)^2 + \left(\frac{u^2 \sin^2 \theta}{2g}\right)^2} \]
\[ = \sqrt{\left(\frac{u^2 \sin \theta \cos \theta}{g}\right)^2 + \left(\frac{u^2 \sin^2 \theta}{2g}\right)^2} \]

2. Factor out \( \left(\frac{u^2}{g}\right)^2 \):
\[ S = \frac{u^2}{g} \sqrt{\sin^2 \theta \cos^2 \theta + \frac{1}{4} \sin^4 \theta} \]

### Step 5: Calculate the Time Taken
The time taken to reach the highest point is half of the total time of flight. The time to reach the maximum height is given by:
\[ t = \frac{u \sin \theta}{g} \]

### Step 6: Calculate the Average Velocity
The average velocity \( V_{avg} \) is given by:
\[ V_{avg} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{S}{t} \]
Substituting the values we calculated:
\[ V_{avg} = \frac{\frac{u^2}{g} \sqrt{\sin^2 \theta \cos^2 \theta + \frac{1}{4} \sin^4 \theta}}{\frac{u \sin \theta}{g}} \]
\[ = \frac{u}{\sin \theta} \sqrt{\sin^2 \theta \cos^2 \theta + \frac{1}{4} \sin^4 \theta} \]

### Final Expression
After simplifying, we find:
\[ V_{avg} = \frac{u}{2} \sqrt{3 \cos^2 \theta + 1} \]