A particle is projected with a speed u at angle `theta` with the horizontal. Consider a small part of its path ner the highest position and take it approximately to be a circular arc. What is the rdius of this circle? This radius is called the adius of curvature of the curve at the point.

Explanation

To find the radius of curvature of the particle's path at the highest point of its trajectory, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components of Motion**:
- The particle is projected with an initial speed \( u \) at an angle \( \theta \) with the horizontal.
- At the highest point of its trajectory, the vertical component of the velocity is zero, and the horizontal component remains \( u \cos \theta \).

2. **Determine the Velocity at the Highest Point**:
- The velocity \( v \) at the highest point is given by:
\[ v = u \cos \theta \]

3. **Understand the Forces Acting on the Particle**:
- At the highest point, the only force acting on the particle is its weight \( mg \) (where \( m \) is the mass of the particle and \( g \) is the acceleration due to gravity).
- The centripetal force required to keep the particle moving in a circular path is provided by this weight.

4. **Set Up the Equation for Centripetal Force**:
- The centripetal force \( F_c \) can be expressed as:
\[ F_c = \frac{mv^2}{r} \]
- At the highest point, the centripetal force is equal to the gravitational force:
\[ mg = \frac{mv^2}{r} \]
- Here, \( r \) is the radius of curvature we want to find.

5. **Substitute the Velocity into the Equation**:
- Substitute \( v = u \cos \theta \) into the centripetal force equation:
\[ mg = \frac{m(u \cos \theta)^2}{r} \]
- Simplifying this gives:
\[ mg = \frac{m u^2 \cos^2 \theta}{r} \]

6. **Solve for the Radius of Curvature \( r \)**:
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g = \frac{u^2 \cos^2 \theta}{r} \]
- Rearranging the equation to solve for \( r \):
\[ r = \frac{u^2 \cos^2 \theta}{g} \]

7. **Final Result**:
- The radius of curvature at the highest point of the trajectory is:
\[ r = \frac{u^2 \cos^2 \theta}{g} \]