Suppose you have an object that is at height h and slides down a length l. How long does it take to slide this distance?
From my understanding, on an inclined plane, $s=\frac{1}{2}sin(\theta)\alpha t^2$, Hence, $l=\frac{1}{2}sin(\theta)\alpha t^2$ where $\theta$ is the angle of the plane and $\alpha$ is the downward acceleration.
The answer provided is $\frac{l}{4\sqrt{h}}$ But I do not know how to get to this expression.
Taking a coordinate system with the $x$-axis parallel to the slope, you get according to Newton's second law:
$$mx^{\prime \prime}(t) = m \alpha \sin \theta$$ where $\alpha$ is the gravity.
And therefore $x(t) = \frac{1}{2} \alpha \sin \theta t^2$ if the object starts at time $t=0$ at the origin of the coordinate system with no velocity. You have $\frac{h}{l} = \sin \theta$.
We're looking for $T$ such that $x(T) = l$ which means
$$l= \frac{1}{2} \alpha \frac{h}{l} T^2$$ and finally
$$T = l \sqrt{\frac{2}{\alpha h}}$$
...and I don't understand where the expression you were given is coming from. By the way, if I understood correctly the question, if $\theta = \pi/2$ you get $l=h$ and the expression I got is coherent with a free falling motion.