A body is projected with a speed u at a angle `theta` with the horizonatl. The radius of curvature of the trajectory, when it makes an angle `((theta)/(2))` with the horizontal is (g-acceleration due to gravity)

Options

A. `(u^(2) cos^(2)theta sec^(3)((theta)/(2)))/(sqrt(3)g)`
B. `(u^(2) cos^(2)theta sec^(3)((theta)/(2)))/(2g)`
C. `(2u^(2) cos^(2)theta sec^(2)((theta)/(2)))/(g)`
D. `(u^(2) cos^(2)theta sec^(3)((theta)/(2)))/(g)`

Answer

D

Explanation

Let velocity of projectile is v at an angle `(theta)/(2)` with horizontal
`therefore v cos " "(theta)/(2) =u cos theta`
or `" "v=(u cos theta)/(cos""(theta)/(2))`

As horizontal component remains same. Also, centripetal force is provided by the component of weight.
So, `(mv^(2))/(r) = mg cos ""(theta)/(2)`
Hence, radius of curvature of path.
`rArr" "r=(v^(2))/(g cos ""(theta)/(2))`
`(u^(2) cos^(2) theta)/(((cos ""(theta)/(2))^(2))/(g cos ""(theta)/(2))) rArr r = (u^(2) cos^(2) theta cdot sec^(3) ((theta)/(2)))/(g)`