A ball suspended by a thread, swings in a vertical plane so that its acceleration in the extreme position and lowest position are equal in magnitude. Angle θ of thread deflection in the extreme position will be:

Step-by-step solution

a A = g s i n θ (only tangential) a B = v 2 l (only radial) K . E . + P . E . = 1 2 m 0 2 + m g l 1 - cos ⁡ θ = 1 2 m v 2 v 2 = 2 g l 1 - cos ⁡ θ ...........(i) ∴ a B = v 2 l = 2 g ( 1 - c o s θ ) Since, a A = a B ∴ g s i n θ = 2 g 1 - c o s θ ⟹ 2 sin ⁡ θ 2 cos ⁡ θ 2 = 2 × 2 sin 2 ⁡ θ 2 ⟹ tan ⁡ θ 2 = 1 2 ⟹ θ = 2 tan - 1 ⁡ 1 2