A ball suspended by a thread swing in a vertical plane so that its acceleration values in the extreme and the lowest position are equal. Find the thread deflection angle in the extreme position.Given ${\cos }^{-1}\frac{2}{\sqrt{5}}=26.{37}^0 and {\tan }^{-1}\frac{1}{2}=26.{57}^0$.

Suppose $\theta$ is the required angle. At extreme position the velocity of the ball is zero, thus normal acceleration $a_c=\frac{v^2}{\mathcal{l}}=0$, and tangential $a_t=g \sin \theta$. At mean position the velocity of the ball $v=\sqrt{2g\left(\mathcal{l}-\mathcal{l} \cos \theta \right)}$.

The normal acceleration at this position 

$$\begin{gathered} a_c=\frac{v^2}{\mathcal{l}} \\ =\frac{2g\left(\mathcal{l}-\mathcal{l} \cos \theta \right)}{\mathcal{l}} \\ =2g\left(1-\cos \theta \right) \end{gathered}$$

and tangential acceleration

                              $a_t=g \sin \theta$

 Thus total acceleration at mean position

$a = 2g (1–\cos \theta )$

  According to given condition, we have 

$g \sin \theta = 2g (1–\cos \theta )$

 After solving, we get $\theta = 53°$