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+3cos^2theta))`
C. `u/2(sqrt(2+cos^2theta))`
D. `u/2 (sqrt(1+cos^2theta))`

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 components of projectile motion
When a projectile is launched with an initial speed \( u \) at an angle \( \theta \) from the horizontal, it has two components of velocity:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

### Step 2: Determine the height at the highest point
The maximum height \( h \) reached by the projectile can be calculated using the formula:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
where \( g \) is the acceleration due to gravity.

### Step 3: Determine the horizontal distance at the highest point
At the highest point, the horizontal distance covered is half of the total range \( R \). The range \( R \) can be calculated using the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Thus, the horizontal distance at the highest point is:
\[ \text{Distance} = \frac{R}{2} = \frac{u^2 \sin 2\theta}{2g} \]

### Step 4: Calculate the displacement from the starting point to the highest point
The displacement \( d \) from the starting point to the highest point can be found using the Pythagorean theorem:
\[ d = \sqrt{h^2 + \left(\frac{R}{2}\right)^2} \]

Substituting the values of \( h \) and \( R \):
\[ d = \sqrt{\left(\frac{u^2 \sin^2 \theta}{2g}\right)^2 + \left(\frac{u^2 \sin 2\theta}{2g}\right)^2} \]

### Step 5: Calculate the time taken to reach the highest point
The time \( t \) taken to reach the highest point is given by:
\[ t = \frac{u \sin \theta}{g} \]

### Step 6: Calculate the average velocity
The average velocity \( v_{avg} \) is defined as the total displacement divided by the total time taken:
\[ v_{avg} = \frac{d}{t} \]

Substituting the values of \( d \) and \( t \):
\[ v_{avg} = \frac{\sqrt{\left(\frac{u^2 \sin^2 \theta}{2g}\right)^2 + \left(\frac{u^2 \sin 2\theta}{2g}\right)^2}}{\frac{u \sin \theta}{g}} \]

### Step 7: Simplify the expression
After simplifying the expression, we find:
\[ v_{avg} = \frac{u}{2} \sqrt{\sin^2 \theta + 3 \cos^2 \theta} \]

### Final Result
Thus, the average velocity of the particle in projectile motion between its starting point and the highest point of its trajectory is:
\[ v_{avg} = \frac{u}{2} \sqrt{1 + 3 \cos^2 \theta} \]

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