Physics - NLM Question with Solution | TestHub

Let the ball accelerate up with $a_1$ and the rod down with $a_2$. The pulley constraint gives $a_2=2a_1$.

Ball: $T-mg=m{a}_1$. Rod: $mg-\frac{T}{2}=m{a}_2$.

Solving: $a_1=\frac{g}{5}$, $a_2=\frac{2g}{5}$.

Relative acceleration of ball with respect to rod: $a_{12}=\frac{g}{5}+\frac{2g}{5}=\frac{3g}{5}$. The ball starts at distance $l$ from the upper end.

$l=\frac{1}{2}\cdot \frac{3g}{5}{t}^2\Rightarrow t=\sqrt{\frac{10l}{3g}}$.