A

Step 1: At the top of the trajectory of the projectile, the vertical component of velocity is zero. The only component of velocity at this point is horizontal.
Step 2: The centripetal acceleration at the highest point can be expressed in terms of the radius of curvature (R) and the horizontal velocity (v): \[ a_c = \frac{v^2}{R} \]
Step 3: The horizontal velocity can be expressed as: \[ v = u \cos(\theta) \]
Step 4: The gravitational acceleration is acting downward, which provides the centripetal acceleration needed for circular motion: \[ g = \frac{v^2}{R} \]
Combining the equations gives: \[ R = \frac{v^2}{g} = \frac{(u \cos(\theta))^2}{g} = \frac{u^2 \cos^2(\theta)}{g} \]
Therefore, the radius of curvature at the highest point of the trajectory is \( \frac{u^2 \cos^2(\theta)}{g} \).
Thus, the answer is A.