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 $\frac{\theta }{2}$ with the horizontal ?

Answer 1

Let 'v' the velocity at the point where it makes an angle $\frac{\theta }{2}$ with horizontal.

The horizontal component remains unchanged.

So, $vcos\left(\frac{\theta }{2}\right)=ucos\theta$

$\Rightarrow v=\frac{ucos\theta }{cos\left(\frac{\theta }{2}\right)}...\left(i\right)$

From figure, $mgcos\left(\frac{\theta }{2}\right)=\frac{m{v}^2}{r}$

$\Rightarrow r=\frac{v^2}{gcos\left(\frac{\theta }{2}\right)}$

Putting the value of 'v' from equation (i)

$r=\frac{v^{2}co{s}^2\theta }{gco{s}^2\left(\frac{\theta }{2}\right)}$