Why isn't centripetal acceleration considered in this problem 
Fairgoers ride a Ferris wheel with a radius of 6.00 m. The wheel completes one 
  revolution every 30.0 s. If a rider accidentally drops a stuffed animal at the top of the wheel, where does it land relative to the base of the ride? (Note: The bottom of the wheel is 1.75 m above the ground.)

Solution: Determine tangential velocity, which is velocity in the horizontal direction = 1.26 m/s
Determine time it takes for bear to fall using $\Delta Y=gt^2/2 = 1.67$ s. Use this information to find $\Delta X = 2.11$ m.
My question is, why isn't the vertical acceleration equal to $g+Ca$?
 A: According to the problem, the stuffed animal has been dropped. That means the only force on it is gravity, so its acceleration after being dropped is $g$. It doesn't "remember" being dropped from a Ferris wheel - it only knows its current position and velocity; nothing else.
A: You can't add centripetal force and gravity, because they aren't the same type of concept.
Gravity, or tension in a string, or friction between a tire and the road are real sources of force.  They are there because of some physical structure and/or interaction.  
Centripetal force is a calculation of the force that is needed to achieve some required circular motion.  The calculation is the first step in determining whether the required motion will even take place.
For example:  You are driving a 1000 kg NASCAR Sprint Cup car, at 240 km/hr into an unbanked curve with a radius of 500 meters.  Being an accomplished multi-tasker, you do some unit conversions, punch a few numbers into the centripetal force formula and come up with$$F_c=\frac{1000\times \left (\frac{240000}{3600}\right)^2}{500}=8889\text{ Newtons}$$ But this doesn't create the force!  And you need one, badly, and right now!
Gravity acts straight down, and the reaction force of the road acts straight up;  neither of these are acting in the direction of the needed centripetal force.  Pushing the accelerator or brake won't help;  these create friction forces acting forward or backward, not inward.  You try turning the wheel slightly, to add a sideways friction force (tires on road) that may be  enough to give you what you want.  Unfortunately, the car in front of you has blown its engine and dumped all its oil on the track.  No friction;  no force to supply the needed centripetal force; no moving in a circle. 
After moving in a straight line for a few seconds, you do find a force that will, one hopes, do the job:  the outside wall will press inward with enough reaction force
to make you travel in a slightly larger circle, and also add friction to slow you down, which also reduces the centripetal force needed.
In the case of the stuffed animal, while on the ride, the only accelerated motion is circular;  you could calculate the centripetal force needed to accomplish this.  Then you could check the various forces acting on the toy:  gravity, radial tension/thrust from the spokes of the Ferris wheel, and tangential force from the bending of the spokes.  You might then decide that the toy was too massive;  that the structure of the wheel could not supply the needed force.
A: 
"My question is, why isn't the vertical acceleration equal to $g+Ca$?"    

I find a very nice explanation on internet.
You are missing some important facts of the situation involved.
Fact 1: There are only two forces acting on stuffed animal throughout its ride these are :
(a) Normal force $N$ and
 (b) The gravity $mg$.
Fact 2: The total force acting on the animal is the resultant of $\vec N$ and $\vec{mg}$. Since the animal is revolving this resultant will be equal to the centripetal force required for the given rotational motion at tangential speed 1.26 m/s i.e $\vec N+\vec {mg}=\vec F_c$ where $\vec F_c$ is the required centripetal force.  

image 1 http://www.physcs.com/wp-content/uploads/2013/01/physics_ferris_wheel_2.png 
In position (1) the magnitude  of $F_c$ is $|mg-N_1|$ in downward direction and in position (2) $F_c$ is in upward direction having magnitude $|N_2-mg|$
 


Eventually the animal is dropped at position (1) 

After the very moment the animal is dropped the Normal force acting on it disappears.
i.e $N$ becomes $0$ leaving the resultant force $mg$ in magnitude So the vertical acceleration is $g$ not $g+Ca$.
