Solution :
Given :
Angle q = angle between the thread supporting the ball with the vertical.
Let mass [tex]$m_1 >>>m_2$[/tex].
Then [tex]$m_1+m_2=m_1$[/tex]
In this case, acceleration can be found out by applying Newton's law of motion.
Thus,
Acceleration, [tex]$a=\frac{m_1}{m_1+m_2}. g$[/tex]
[tex]$a=\frac{m_1}{m_1}. g$[/tex]
[tex]$a=g$[/tex]
Therefore, [tex]$\tan \theta =\frac{a}{g}$[/tex]
or [tex]$\tan \theta =\frac{a}{a}$[/tex]
or [tex]$\tan \theta =1$[/tex]
[tex]$\theta = \tan ^{-1}(1)$[/tex]
[tex]$\theta = 45^\circ$[/tex]
Therefore the largest angle q is [tex]$\theta = 45^\circ$[/tex]
