How to measure time taken by object descending quadratic curve

Solution 1:

$ds = \sqrt{1+y'(x)^2}\, dx$ is the distance element.

$v =\sqrt{2\,g\,y(x)}$ is the velocity. It comes from the kinetic energy equaling the potential energy. $\frac{mv^2}{2} = mgh$

But $y(0)=\frac{36}{9} = 4$ so the initial velocity is not zero.

$y(6) = 0$ the velocity at the bottom is zero.This is incorrect. If the velocity is zero it has stopped moving so time will accumulate indefinitely.

Replace $y(x)$ in the denominator with $y(0)-y(x)$.

$\Delta h = y(0)-y(x)$