A particle is projected with initial speed u at an angle `theta` above the horizontal . Let A be the point of projection , B the point where velocity makes an angle `theta //2` above the horizontal and C the highest point of the trajectory .
Radius of curvature of the trajectory at point B is

Options

A. `(u^(2) cos ^(2) theta)/(2g cos^(3) (theta)/(2))`
B. `(u^(2) cos ^(2) theta)/(g cos ^(3) (theta)/(2))`
C. `(2u^(2) cos ^(2) theta)/(g cos ^(3) (theta)/(2))`
D. `(u^(2) cos ^(3) theta)/(g cos ^(2) (theta)/(2))`

Explanation

To find the radius of curvature of the trajectory at point B, we can follow these steps:

### Step 1: Understand the motion of the projectile
A particle is projected with an initial speed \( u \) at an angle \( \theta \) above the horizontal. The motion can be analyzed in two dimensions: horizontal and vertical.

### Step 2: Determine the velocity at point B
At point B, the velocity makes an angle \( \frac{\theta}{2} \) with the horizontal. The horizontal and vertical components of the velocity can be expressed as:
- Horizontal component: \( v_x = u \cos \theta \)
- Vertical component: \( v_y = u \sin \theta - g t \)

At point B, the angle of the velocity is \( \frac{\theta}{2} \), so:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{v_y}{v_x} \]
This gives us the relationship between the vertical and horizontal components of the velocity at point B.

### Step 3: Relate the components of velocity
Using the tangent function:
\[ v_y = v_x \tan\left(\frac{\theta}{2}\right) \]
Substituting \( v_x = u \cos \theta \):
\[ v_y = u \cos \theta \tan\left(\frac{\theta}{2}\right) \]

### Step 4: Find the centripetal acceleration
The centripetal acceleration \( a_c \) is given by:
\[ a_c = \frac{v^2}{R} \]
where \( R \) is the radius of curvature and \( v \) is the speed at point B. The speed \( v \) can be calculated as:
\[ v = \sqrt{v_x^2 + v_y^2} \]

### Step 5: Calculate the forces acting on the particle
At point B, the forces acting on the projectile are the gravitational force \( mg \) acting downwards and the normal force providing the centripetal force. The component of the gravitational force acting towards the center of curvature is:
\[ F_c = mg \cos\left(\frac{\theta}{2}\right) \]

### Step 6: Set up the equation for the radius of curvature
Using the centripetal force equation:
\[ \frac{mv^2}{R} = mg \cos\left(\frac{\theta}{2}\right) \]
This simplifies to:
\[ R = \frac{v^2}{g \cos\left(\frac{\theta}{2}\right)} \]

### Step 7: Substitute the expression for speed
Now we need to substitute \( v \) back into the equation. From step 4:
\[ v = \sqrt{(u \cos \theta)^2 + (u \cos \theta \tan\left(\frac{\theta}{2}\right))^2} \]
This simplifies to:
\[ v = u \cos \theta \sqrt{1 + \tan^2\left(\frac{\theta}{2}\right)} = u \cos \theta \sec\left(\frac{\theta}{2}\right) \]

### Step 8: Final expression for the radius of curvature
Substituting \( v \) into the radius of curvature equation:
\[ R = \frac{(u \cos \theta \sec\left(\frac{\theta}{2}\right))^2}{g \cos\left(\frac{\theta}{2}\right)} \]
This simplifies to:
\[ R = \frac{u^2 \cos^2 \theta}{g \cos^3\left(\frac{\theta}{2}\right)} \]

### Final Answer
Thus, the radius of curvature of the trajectory at point B is:
\[ R = \frac{u^2 \cos^2 \theta}{g \cos^3\left(\frac{\theta}{2}\right)} \]