Question:
What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where the particle velocity makes an angle $0 / 2$ with the horizontal?
Solution:
Since acceleration in horizontal direction is zero
$u \cos \theta=\operatorname{vcos}\left(\frac{\theta}{2}\right)$
$v=\frac{u \cos \theta}{\cos _{\frac{\theta}{2}}^{\theta}}$
$\mathrm{a}_{\text {radial }}=\frac{V^{2}}{R}$
$R=\frac{\frac{w^{2} \cos ^{2} \theta}{\cos ^{2} \theta / 2}}{g \cos \theta / 2}$
$R=\frac{u^{2} \cos ^{2} \theta}{g \cos ^{3}(\theta / 2)}$