The question asks which block hits the floor first. One solution argues that the first block hits the ground first. This is because the the slope of that incline is steeper, so the block reaches the bottom of the incline faster.
This problem can be simplified by comparing two ramps of different slopes.
At the bottom of both ramps, the object has the same speed. The object with the steeper slope, however, reaches that speed in a shorter time.
It's important to note that the second ramp occupies a smaller area then ramp one in the rectangle formed by $xh$.
From basic kinematics, the time needed to traverse the horizontal distance $x$ for the first ramp is:
$$t = \sqrt{\frac{s_1}{2g sin \theta}}$$
The time to travel the same horizontal distance $ x$ for the second ramp has two parts: The time to traverse the second ramp:
$$t_1 = \sqrt{\frac{s_2}{2g sin \phi}}$$
and the time to traverse the remaining horizontal distance b:
$$t_2 = \frac{b}{v_f}$$ where $v_f$ is the velocity at the end of the ramp. If $t' = t_1 + t_2$, then the ratio of the two times for each ramp is:
$$\frac{ t'} t = \frac{(\sqrt{\frac{2 s_2}{gsin\phi}}+\frac{b}{v_f})}{\sqrt{\frac{2 s_1 }{gsin\theta } } }$$
This can be simplified since $s_1 = \frac{h}{sin\theta}$ and $s_2 = \frac{h}{sin\phi}$
and $$v_f = \sqrt{2s_1gsin\theta}$$
This leaves:
$$\frac{t'}{t } = \frac{ sin\theta}{sin\phi} +\frac{bsin\theta}{2h}$$ with $$b = x-hcot\phi$$
Since $tan\theta = \frac{h}{x}$ then,
$$\frac{t'}{t} = \frac{sin\theta}{sin\phi} + \frac{sin\theta(cot\theta - cot\phi)}{2}$$
The ratio of times are now expressed only as a function of the two angles or ramp slopes.
While it's not obvious, if $\phi > \theta$, then t > t' which supports the argument that the path with largest average slope will be the faster path. Also, the longer path may be the faster path. The path length of second ramp is longer than the path of the first ramp.
The graph has $\theta$ fixed at 10° while $\phi$ starts at 10° and increases. There are two competing effects wrt. ramp two:
- The increase in acceleration due to increasing slope.
- The increase in path length due to increasing slope.
These effects are minimized when $\phi - \theta = 45°$
Note that the ramp that occupies the smallest area over the distance x and a height of h will cover the distance x in the least time.
Looking at your diagram, it is clear the curved ramp for block one occupies less area then the second multi-curved ramp. So block 1 arrives first.
This short video illustrates the concept.
https://youtu.be/iogVVja1MYY?si=m1NB5NyYvHP4tvMO