A ball attached to one end of a stringj swings in a vertical plane such that its acceleration at point `A` (exterme posiuion) is eqal to its acceleration at point `B` (mean posirion), The angle `theta` is

Options

A. `cos^(-1)((2)/(5))`
B. `cos^(-1)((4)/(5))`
C. `cos^(-1)((3)/(5))`
D. None of these

Answer

C

Explanation

`a_(A)=gsintheta`
`a_(B)=(v^(2))/(R)=(2gh)/(R)=(2gR(1-costheta))/(R)`
Given, `a_(A)=a_(B)`
:. `sintheta=2-2costheta`
Squaring these two equations we have,
`sin^(2)theta=4+4cos^(2)theta-8costheta`
or `1-cos^(2)theta=4+4cos^(2)theta-8costheta`
On solving this equation, we get
`costheta=(3)/(5)` or `theta=cos^(-1)(3//5)`