\(\therefore \) Acceleration down the plane \( = g\sin \theta \)

Since \(l = 0 + \frac{1}{2}g\sin \theta {t^2}\)

\(\therefore {t^2} = \frac{{2l}}{{g\sin \theta }} = \frac{{2h}}{{g{{\sin }^2}\theta }} \Rightarrow t = \frac{1}{{\sin \theta }}\sqrt {\frac{{2h}}{g}} \)