SOLUTION: Thanks in advance for considering this question. Recently on a desert trip to Utah, I dropped a stone from the ledge of a canyon and timed the return sound of the stone hitting

Question 105204: Thanks in advance for considering this question.
Recently on a desert trip to Utah, I dropped a stone from the ledge of a canyon and timed the return sound of the stone hitting the bottom. Ultimately hoping to calculate the distance to the bottom of the canyon(ignoring certain real world phenomenom such as the whole terminal velocity vs. accelleration due to gravity in a vacuum and the fact that sound doesn't travel in a vacuum thing)I have attempted to solve for the following 3 unkowns algebraically. The total time (TT) is the summation of the time for the rock to hit bottom (TR) plus the time for the sound to return (TS).
TT = TR + TS
Assuming an initial velocity of 0, the distance (D) can be represented by the formula:
D = (.5)(accelleration due to gravity)(TR squared)= (.5)(32f/s/s)(TR squared)
Therefore TR = the square root of the quantity [D/[(.5)(32f/s/s)]]
Sound travels at 1087 ft. per second (ignoring certain environmental factors)and therefore TS = D/1087
TT is known, TR, TS, and D are unknown. We have three equations and 3 unknowns. I started by using TT = TR + TS and replacing TR and TS with the two distance equations i.e. TT = [square root of the quantity D/[(.5)(32f/s/s)]]+[(D/1087)] and trying to solve for D. Can't seem to get er done. Please help with solution and insights (solve for TT in general and I'll be back in the desert throwing rocks down canyons of known depth in no time)!
Thanks
Answer by scott8148(5896) (Show Source):
You can put this solution on YOUR website!
your assumptions and parameters seem good

TR=sqrt(D/16)=(sqrt(D))/4 ... TS=D/1087 ... let x=sqrt(D) ... TT=x/4+x^2/1087

(1/1087)x^2+(1/4)x-TT=0 ... .000919x^2+.25x-TT=0

... remember to square the x to get D

so much for the math, now the reality ... I'm not sure which x value (+ or -) will give the correct D value
...you could try a few known D's and see what happens
...hopefully, it is always the same value

I screwed up the decimal points the first time around ... this should work

2nd edit - the formula works and it is the positive x value that you want
...this is 3-figure accuracy (slide-rule, if you're that old); the question is "can you measure TT that close"?