A particle is projected with a speed u at an angle θ to the...

The horizontal component of velocity remains constant throughout the projectile motion.

Let $v$ be the velocity of the particle when it makes an angle $\theta/2$ with the horizontal. Equating the horizontal components of velocity:

$v \cos(\theta/2) = u \cos\theta$

$v = \dfrac{u \cos\theta}{\cos(\theta/2)}$

The radius of curvature $\rho$ at any point on the trajectory is given by $\rho = \dfrac{v^2}{a_n}$, where $a_n$ is the normal component of acceleration.

The only acceleration acting on the projectile is the acceleration due to gravity $g$, which acts vertically downwards. The component of $g$ perpendicular to the velocity vector is:

$a_n = g \cos(\theta/2)$

Substituting the values of $v$ and $a_n$ into the radius of curvature formula:

$\rho = \dfrac{\left( \dfrac{u \cos\theta}{\cos(\theta/2)} \right)^2}{g \cos(\theta/2)}$

$\rho = \dfrac{u^2 \cos^2\theta}{g \cos^3(\theta/2)}$