Seriously, man, watch the 1/8 speed copy. It's pretty clear. The impact and beginning of bounce for both objects occurs between frames. The first image you posted is just before impact of either, and the one in your last post is one frame before that. None of the actual impacts of either object in any of the drops can be seen in a still frame. Would need probability of many more drops or a much higher speed camera to get that.
JohnKSa hit the physics pretty well in post #41. Acceleration is caused by an imbalance in forces, in the case of the bolt and ear plug, the two dominant forces are gravity and drag. In the atmosphere, the acceleration of neither the bolt, nor the earplug will remain 32.2 ft/s^2 after drag forces start to build.
The inverse square law says that the gravitational force on an object is related to its mass (and the mass of the earth). If you're just looking for the acceleration of that object in a vacuum, the mass term is cancelled out when you combine this equation with F=MA. If your looking at acceleration in the atmosphere you can't assume constant acceleration and you have to do a force analysis for each object that will be dependent on its drag and mass. The acceleration of the bolt and the ear plugs will be different, but you probably won't be able to see the difference by dropping them 7 feet be hand, since they won't be going that fast when they hit the ground, and drag force is proportional to the square of velocity. If you drop them from significantly higher you'll see a difference even though neither object hits is terminal velocity.
In the scope of human knowledge on physics, I have a very cursory understanding. Never took it in school (9th grade drop out), never researched it beyond what I needed to know. Nonetheless, I do understand that drag force is basically velocity^2/drag coefficient, accounting for air density. So, two objects with the same drag coefficient but of very different densities have very different terminal velocities. That is elementary. The more involved aspect of acceleration under the force of gravity encompasses that drag component; the denser object will overcome greater drag than the more buoyant one, but the drag (friction) becomes exponentially higher as the speed increases.
My contention is that, despite their different respective terminal velocities, the reduction in acceleration from maximum rate to 0 will occur over about the same amount of time. The more dense object will obviously be traveling at a higher speed when the drop in acceleration occurs, so will cover more distance during that time, but the drag forces will also be exponentially higher, so I don't believe the chronology of that change will be much different than for the less dense object. I do not have the math skills to calculate it, and I would certainly love it if someone here who did could take the time to demonstrate it in hard numbers. Even in the absence of that detailed knowledge, I believe that objects in free fall spend the extreme majority of their time between 0 and terminal velocity accelerating at about 32 ft/sec^2, that the reduction in acceleration rate happens in a very short time relative to the time it takes to achieve terminal velocity.
This is why I argue that two very different objects will accelerate at virtually the same rate when falling in earth atmosphere, so long as the distance covered is insufficient to reach or nearly reach terminal velocity, where the reduction in acceleration becomes a very steep curve.
Anyone who has watched my little clip (particularly the slow motion version I posted today) and paid close attention can see that the impact times are so close together that the camera's 45 ms frames missed the actual impacts of both objects all 3 times. This demonstrates that the acceleration rates are very, very close within that 6-1/2 foot drop. Now, if we stretch that distance enough that the ear plug has just reached it's terminal velocity when it hits, I'm certain the bolt would impact noticeably sooner, since it would still be accelerating. I do not, however, know what the required height would be.
