How to determine the depth of a mine shaft

We came across this shaft in the west desert of Utah and decided to drop some rocks down it and see how long it took to hit the bottom.

We determined that a rock falls at 32.2 ft/sec. and increases by that amount every second of free fall until it reaches terminal velocity (which based on the diameter of our rock would be around 170'/second.  So the breakdown of how far it fell per second is below:

1st second:  32.2'
2nd second: 64.4'
3rd second: 96.6'
4th second: 128.8'
5th second: 160'
6th second: 170' (terminal velocity reached)
7th second: 170'
8th second: 170'
9th second: 170'

Total feet the rock fell in 9 second - 1,162' !


Anonymous said...

Your rock-fall timing is slightly off, because of the time it takes for the sound in the impact to return to the surface. For a 1,000 foot shaft, the returning sound would take approximately a full second.

Anonymous said...

If I fell in it wouldn't make any difference, at least to me.

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