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Physics Problem

toxcrusadr

toxcrusadr

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Here's one for you physics nerds to stretch your brain with!


A man drops a rock into a mine shaft. 12 seconds later, he hears the sound of the rock hitting the bottom.

Assume the rock starts from the ground surface at zero velocity, that it hits nothing on the way down, and that air drag is negligible.

How deep is the shaft?

Total time = 12sec = Time for rock to fall + time for sound to return
Speed of sound = 1126 ft/sec
Distance = 1/2at^2 where a is the acceleration of gravity, 9.8 m/s^2

I got lost in the math after this.

Here's a video of someone actually doing it. I was just curious how deep the hole was.

http://www.wimp.com/deepesthole/
 
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For sake of common calculation can we assume the rock was dropped from the level of the hole opening and that's where we listened for the sound too?
 
s is your vertical displacement
v0 is your initial velocity (nothing since it is dropped)
g is gravity
t is travel time

s = (v0 * t) - (1/2 * g * t^2)

s = (0 ft/s * 12s) - (1/2 * 32 ft/s^2 * 12s ^2)

s = 0 - ( 1/2 * 32 * 144) ft

s =-2304 ft

Assuming it's on earth and air resistance is negligible

AKA pretty darn deep! :D


EDIT: This didn't account for the amount of time required for the sound to travel back up the hole, so that would shorten the hole!
 
Can you get the rock back and try it again? That exactly 12 seconds sounds contrived to me.
 
Sounds like when these two guys were walking in a field...

...and they found this deep, dark hole.

The wondered how deep it was and decided to drop a rock down it to see how long it took to hit bottom.

Well, they threw a rock in, waited, and heard nothing.They decided it wasn't big enough and then found an old auto transmission, dragged it to the edge and pushed it in. ..and waited.

All of a sudden a goatl comes running, about a hundred miles an hour, and dives right into the hole.

They look at each other, stunned.

A minute or so later a farmer comes by and says "Either of you guys see my goat?"

The guys again look at each other.

Then one of 'em says "Nah, it's probably running around the field somewhere."

The farmer says, "Not really likely. I had him tied to a transmission."
 
Good one markw!

FreeOrange: accounting for the time it takes the sound to travel back is what makes this a difficult problem! I figured it as:

12 sec = time for rock to fall (Tf) + time for sound to return (Ts)

Tf can be derived as you did using distance = 1/2 a(Tf)^2

Assume a = about 30 ft/sec^2 so everything is in feet

Now, sound travels 1126 ft/sec

so Ts = distance/1126

Substituting this into the initial equation,

12 sec = Tf + Distance/1126

I thought this last equation could be solved for Tf, i.e. Tf = 12 - D/1126

and then plug this into the speed equation D = 1/2aTf^2

giving

Distance = 1/2*(30 ft/sec^2) * (12 sec - D ft/1126 ft/sec)^2

I fell down on the algebra and could not solve this for D. There is probably an easier way to do it.

You can guess at it and plug values in and see if it works out, and I came out with about 10 sec. of fall and 2 sec. of return time, or about 2300 ft.

I hope someone can do this more elegantly.
 
Oh, ain't had anyone ever ask that type of question since 12th grade physics exam. About the like some guy dropped a rock and feather from the leaning tower of pizza and found that terminal velocity regardless of mass was 120 miles per second or something like that. Think it was DiVinci.
 
tox,

Yeah somehow we need to get the solution back to a single variable problem:

t1 = time to fall
t2 = time to hear the impact
Vs = speed of sound
g = gravity
s = depth of hole

12 = t1 + t2 (time observation)

s = 1/2 * g * t1^2 => t1 = sqrt(2*s/g) (reconfiguring the depth forumla to find drop time)

t2 = s/Vs (reconfigure v = s/t to find echo time)

s/Vs + sqrt (2*s/g) = 12 (plug in both of the above equations into the time observation to reduce to the singe variable "s")

2*s/g = (12-s/Vs)^2 (one more manipulation to elminate the sqrt)

This turns into a quadratic which is solvable for s, since everything else is a constant. My head hurts, I'm not sure I can do that one in a text box :D
 
Were you standing, or laying on the ground with your ear to the hole when the rock hit bottom? Also, does a rock dropped in a hole emit sounds that the human ear can't detect when it first hits? Did the weight and velocity of the falling rock displace dirt at the bottom, making the hole deeper, and did that sound vary from when the rock actually stopped?
 
I'm more with whoaru99, except I get that the hole is 1754.113 feet deep.

The rock falls for 10.442 seconds and it takes 1.588 second for the sound to travel back up the hole, 10.442 + 1.588 = 12 seconds.

The three equations are
1. x = x0+v0*tf+0.5*a0*tf^2
2. x = vs*ts
3. t = tf+ts

where x is the hole depth, tf is the time the rock falls, ts is the time it takes for the sound to travel back up the hole, t is the total time, x0 = 0 feet, v0 = 0 feet/sec, a0 = -32.174 ft/sec^2, and vs = -1126 ft/sec (speed of sound).

This leads to a quadratic equation in x.

(x^2)/2 – [(vs+v0)/a0+t]*vs*x+[2*(x0+v0*t)/a0+t^2]*(vs^2)/2 = 0

The solution is

x = vs*[(vs+v0)/a0+t-sqrt{[(vs+v0)/a0+t]^2-2*(x0+v0*t)/a0-t^2}]

If anybody wants the derivation, my head won’t hurt to type it up.

EDITION - PS: as requested by toxcrusadr,

tf = sqrt{[(vs+v0)/a0+t]^2-2*(x0+v0*t)/a0-t^2} - (vs+v0)/a0

ts = t + (vs+v0)/a0 - sqrt{[(vs+v0)/a0+t]^2-2*(x0+v0*t)/a0-t^2}
 
Last edited:
FreeOrange said:
This turns into a quadratic which is solvable for s, since everything else is a constant. My head hurts, I'm not sure I can do that one in a text box :D
Click to expand...

That was my problem, I got to a quadratic and couldn't remember how to solve it.

bentpencil said:
Were you standing, or laying on the ground with your ear to the hole when the rock hit bottom? Also, does a rock dropped in a hole emit sounds that the human ear can't detect when it first hits? Did the weight and velocity of the falling rock displace dirt at the bottom, making the hole deeper, and did that sound vary from when the rock actually stopped?
Click to expand...

You go sit in the corner! :D

aabottom said:
This leads to a quadratic equation in x.

(x^2)/2 – [(vs+v0)/a0+t]*vs*x+[2*(x0+v0*t)/a0+t^2]*(vs^2)/2 = 0

The solution is

x = vs*[(vs+v0)/a0+t-sqrt{[(vs+v0)/a0+t]^2-2*(x0+v0*t)/a0-t^2}]

If anybody wants the derivation, my head won’t hurt to type it up.
Click to expand...

Again with the quadratic! I don't care how you derived it, but if you can solve it for t1 and t2 you can stay after class and clean the erasers!
 
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