What is the radius of curvature of the parabola traced out by the projectile in which a particle is projected with a speed u at an angle `theta` with the horizontal, at a point where the velocity of particle makes an angle `theta//2` with the horizontal.

Answer

A, B, C

Explanation

Let v the velocity at the point where it makes an angle `theta/` with horizontal.
The horizontal component remains unchanged.
So, `v cos (theta/2)=ucostheta`

`rarr v=(u costheta)/(cos(theta/2))`…….i
From ure, `mgcos(theta/2)=(mv^2)/r`
`rarr r=v^2/(gcos(theta/2))`
putting the value of v from equation i.
`r=(v^2cos^2theta)/(g cos^2(theta/2))`