Question

A particle is projected with a speed u at an angle $\theta$ with the horizontal. What is the radius of curvature of the parabola traced out by the projectile at a point where the particle velocity makes an angle $\frac{\theta }{2}$ with the horizontal.

Answer 1

Let v be the velocity at the desired point. Horizontal component of velocity remains unchanged. Hence,
$v\cos \frac{\theta }{2}=u\cos \theta$
$v=\frac{u\cos \theta }{\cos \frac{\theta }{2}}$ ...(i)
Radial acceleration is the component of acceleration perpendicular to velocity or
$a_n=g\cos \left(\frac{\theta }{2}\right)$
$\therefore \frac{v^2}{R}=g\cos \left(\frac{\theta }{2}\right)$
Substituting value of v from Eq. (i) in Eq. (ii), we have radius of curvature
$R=\frac{\left[\frac{u\cos \theta }{\cos \left(\frac{\theta }{2}\right)}\right]}{g\cos \left(\frac{\theta }{2}\right)}=\frac{u^2{\cos }^2\theta }{g{\cos }^3\left(\frac{\theta }{2}\right)}$