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?

Answer 1

The radius of curvature of the parabola is $\begin{equation}\frac{u^{2} \cos ^{2} \theta}{g \cos ^{2} \theta / 2}$

Explanation:

Step 1:

Now, We know that in case of the Projectile motion, Horizontal velocity remains the same throughout the entire motion.

Horizontal velocity at first = u cos θ

At Highest Point, Resultant velocity = u cos θ [Since, vertical will be zero.]

Step 2:

Now, If the velocity vector is making an angle θ/2 with the horizontal at some point in its trajectory, then

Horizontal component =$\begin{equation}=v \cos \frac{\theta}{2}$

Then,

$\begin{equation}\begin{aligned}&v \cos \frac{\theta}{2}=u \cos \theta\\&v=\frac{\cos \frac{\theta}{2}}{\cos \theta}\\&\frac{m v^{2}}{r}=m g\\&m v^{2}=r m g\\&\frac{m v^{2}}{m g}=r\\&r=\frac{v^{2}}{g}\\&r=\frac{u^{2} \cos ^{2} \theta}{g \cos ^{2} \theta / 2}\end{aligned}$