#### 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 θ/2 with the horizontal?

#### Solution

Let *u* be the initial velocity and *v* be the velocity at the point where it makes an angle \[\frac{\theta}{2}\] with the horizontal component. It is given that the horizontal component remains unchanged.

Therefore, we get :

\[v \cos \left( \frac{\theta}{2} \right) = u cos\theta\]

\[\Rightarrow v = \frac{u\cos\theta}{\cos\frac{\theta}{2}} . . . \left( i \right) \]

\[mg\cos\frac{\theta}{2} = \frac{m v^2}{r} . . . \left( ii \right)\]

\[ \Rightarrow r = \frac{v^2}{g\cos\frac{\theta}{2}}\]

On substituting the value of *v* from equation (i), we get :

\[r = \frac{u^2 \cos^2 \theta}{g \cos^2 \frac{\theta}{2}}\]