$$t = \frac{\pi}{2} \sqrt{\frac{l}{\mu g}} \quad \text{for} \quad v \leq \sqrt{\mu g l}, \quad t = \frac{v'}{\mu g} + \sqrt{\frac{l}{\mu g}} \arccos \frac{v'}{v} \quad \text{for} \quad v > \sqrt{\mu g l}, \quad \text{where}$$

$$v' = \sqrt{v^2 - \mu g l}.$$