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

At the highest point of the trajectory, the velocity of the particle is purely horizontal and is given by $v = u \cos\theta$.

The acceleration acting on the particle is the acceleration due to gravity $g$, which is directed vertically downwards.

Since the velocity is perpendicular to the acceleration at this point, the entire acceleration acts as the centripetal acceleration.

Using the formula for radius of curvature $R = \dfrac{v^2}{a_{\perp}}$:

$R = \dfrac{(u \cos\theta)^2}{g}$

$R = \dfrac{u^2 \cos^2\theta}{g}$