A particle is projected with a speed u at angle `theta` with the horizontal. Consider a small part of its path near the highest position and take it approximately to be a circular arc. What is the radius of this circle? This radius is called the adius of curvature of the curve at the point.

Answer

A, B, C

Explanation

`a=(v^(2))/(R) implies R=(v^(2))/(a)=((ucostheta)^(2))/(g)`