A ball suspended by a thread swings in a vertical plane so that its acceleration in the extreme position and lowest position are equal. The angle  of thread deflection in the extreme position will be - 2

Explanation

The ball has only normal acceleration at the lowest position and only tangential acceleration at any of the extreme position. Let v be the speed of the ball at its lowest position and l be the length of the thread, then according to the problem
`v^2/l=g sin alpha` (1)
wehre `alpha` is the maximum deflection angle
From Newton's law in projection form: `F_t=mw_t`
`-mg sin theta=mv(dv)/(l d theta)`
or, `-gl sin theta d theta=v dv`
On integrating both the sides within their limits.
`-gl underset(0) overset(alpha)int sin theta d theta=underset(v)overset(0)int v dv`
or, `v^2=2gl (1-cos alpha)` (2)
Note: Eq. (2) can easily be obtained by the conservation of mechanical energy of the ball in the uniform field of gravity.
From Eqs. (1) and (2) with `theta=alpha`
`2gl(1-cos alpha)=lg cos alpha`
or `cos alpha=2/3` so, `alpha=53^@`