What is the radius of curvature of the parabola traced out by the projectile in which a particle is projected by a speed u at an angle 'theta' with the horizontal, at a point where the particle velocity makes and angle `theta/2` with the horizontal?

Explanation

To find the radius of curvature of the parabola traced out by a projectile at a point where the particle's velocity makes an angle of θ/2 with the horizontal, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Velocity Components:**
The initial velocity \( u \) is projected at an angle \( \theta \) with the horizontal. The components of the initial velocity are:
\[ u_x = u \cos \theta \quad \text{(horizontal component)} \]
\[ u_y = u \sin \theta \quad \text{(vertical component)} \]

2. **Determine the Velocity Components at the Point of Interest:**
At the point where the velocity makes an angle \( \theta/2 \) with the horizontal, we denote the velocity at this point as \( v \). The components of \( v \) can be expressed as:
\[ v_x = v \cos \left(\frac{\theta}{2}\right) \quad \text{(horizontal component)} \]
\[ v_y = v \sin \left(\frac{\theta}{2}\right) \quad \text{(vertical component)} \]

3. **Use the Horizontal Velocity Conservation:**
Since there is no horizontal acceleration, the horizontal component of the velocity remains constant:
\[ u_x = v_x \implies u \cos \theta = v \cos \left(\frac{\theta}{2}\right) \]
From this, we can solve for \( v \):
\[ v = \frac{u \cos \theta}{\cos \left(\frac{\theta}{2}\right)} \tag{1} \]

4. **Determine the Radial Acceleration:**
The radial acceleration \( a_r \) at the point of interest is given by the component of gravitational force acting perpendicular to the velocity. The vertical component of gravitational force acting at this angle is:
\[ a_r = g \cos \left(\frac{\theta}{2}\right) \]

5. **Relate Radial Acceleration to Radius of Curvature:**
The radius of curvature \( R \) is related to the velocity and radial acceleration by the formula:
\[ a_r = \frac{v^2}{R} \implies R = \frac{v^2}{a_r} \]
Substituting the expressions for \( v \) and \( a_r \):
\[ R = \frac{v^2}{g \cos \left(\frac{\theta}{2}\right)} \]

6. **Substituting for \( v \):**
Now substitute the expression for \( v \) from equation (1):
\[ R = \frac{\left(\frac{u \cos \theta}{\cos \left(\frac{\theta}{2}\right)}\right)^2}{g \cos \left(\frac{\theta}{2}\right)} \]
Simplifying this gives:
\[ R = \frac{u^2 \cos^2 \theta}{g \cos^3 \left(\frac{\theta}{2}\right)} \]

### Final Result:
The radius of curvature \( R \) of the parabola traced out by the projectile at the specified point is:
\[ R = \frac{u^2 \cos^2 \theta}{g \cos^3 \left(\frac{\theta}{2}\right)} \]