How to calculate the time it takes for an object to fall on a curved path? Imagine an object falling along a quarter circle of height $h$ meters ($h$ would also be the radius) and gravity $g$ meters per second squared. How long will it take for the object to travel the distance of the curve in terms of $h$ and $g$? (no friction, no rolling, the object is negligibly small compared to the quarter circle, and the object is stationary until $t = 0$.)

I was able to determine the following answer:
$${t(h, g)=\sqrt{\frac{h}{g}}\cdot1.854074...}$$
The only issue is that I have no idea where the magic 1.854074... number comes from. This was solved with brute force calculation and curve fitting, but I'm sure a proper solution to this problem will better explain the value.
 A: As John Forkosh said in the comments this can be done in a similar way to the brachistochrone problem problem. But in this case I think this overly complicates things. Instead we can use the fact that we are on a circle to find the speed. From conservation of energy we have:
$$\frac{1}{2}mv^2=mgy$$
where $y$ is measured from the top of the circle downwards. Putting this in terms of angular velocity:
$$\frac{1}{2}mh^2 \dot \theta^2=mgh \sin(\theta)$$
$$\int \frac{1}{\sqrt{2\sin(\theta)}}d\theta=\sqrt{\frac{g}{h}}\int dt$$
A: My answer is quite late, but I hope someone will find it helpful.
In fact, we can use Newtonian mech to solve this problem.
We first decompose the gravitational force into two components--tangential and centripetal. The centripetal force does not change the magnitude of the velocity, so let's ignore it.
Now we look at the tangential component of the gravity.
$$ma_t=mg\cdot\cos\theta$$
we know that $a_t=R\ddot\theta$, so
$$R\ddot\theta=g\cos\theta$$
$$\ddot\theta=\frac g l \cos\theta$$
(The equation above is similar to the equation of a SHM pendulum)
Now what we need to do is to solve this ODE. Since this is not easy to do, I will leave the answer like that.
A: I am generally averse to assist with anything resembling a homework problem but, in this instance, I think answering would be appropriate.
The question being asked is technically a wholly other question in disguise: the information presented defines a pendulum and you are being asked to find HALF the period.
The formula to find the complete period is:
$$T ≈ 2\pi \sqrt\frac{L}{g}$$
Just remember that, in this instance, L and h are the same and you only want HALF the complete period.
