Realistic acceleration (in g's) check?

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Dropping an object will take approximately 0.102 seconds to reach a speed of 1 m/s, which aligns with classical physics equations. When the object is stopped after reaching 1 m/s, a stopping distance of 1 cm results in a net acceleration of about 5g, while a 3 mm stopping distance increases this to around 17g. This relationship can be derived by equating kinetic energy and work, demonstrating that halving the stopping distance requires doubling the force for a fixed kinetic energy. The discussion confirms the calculations are accurate and highlights the inverse proportion between stopping distance and required force. Understanding these principles is essential for realistic assessments of acceleration in physics.
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Folks - doing some research and just needed a reality check on acceleration - let me know if this makes sense:

If I drop an object, it should take roughly 5 cm (.102 sec) to reach 1 m/s, correct? I'm using the classical physics equations for this (please do a quick verification on this).

Then, if I stop the object after it reaches 1 m/s, and the stopping distance is 1 cm, then I'm getting a NET accelleration of ROUGHLY 5g. Then if that stopping distance decreases to 3 mm it jumps to around 17g. Again, this is a net acceleration.

Just want to verify, I know this is easy, but figured i might as well as post it since I'm rusty at this.
 
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Yes, this is correct.

You can derive the relationship based on equating kinetic energy and work. Halve the distance and you double the required force for a fixed kinetic energy. It's a simple inverse proportion.
 

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