Timed freefall + sound to respond from cave

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To determine the depth of the abyss, the total time of 9 seconds must be split into the time it takes for the rock to fall (x) and the time for the sound to travel back up (9 - x). The equations for distance down and distance up must be equal, allowing for the elimination of one variable. After calculations, two possible depths of 303 meters and 319 meters were proposed, with a suggestion to verify by checking which depth yields a total time of 9 seconds when the fall and sound travel times are calculated. The discussion emphasizes the importance of correctly applying physics formulas to solve the problem.
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Homework Statement


"an ex-physics student is exploring a cave and he comes to a deep abyss. since he has a stopwatch, he decides to determine how deep the abyss is. He remembers that sound travels at a constant velocity of 345 m/s he drops a rock and times how long it is until he hears it hit the bottom, which is 9.00 s. This is the time it takes to fall to the bottom and for the sound to travel back up to the person. Calculate how deep the abyss is in meters.

TWO hints:
1) you msut first determine the part of the 9.00s for the fall and for the sound to travel back up. let x represent the time for the fall, and then (9.00- x) represents the time for the sound to travel back up the abyss.

use your formulas so that you can eliminate all vars except for X"


Homework Equations



I just started in a basic physics class for high school and this is a homework problem that I am absolutely stumped by. I don't understand how to calculate for the time it takes to get down or the time to get back up, cause wouldn't it be like trying to figure out what X and Y are in a formula like 10=xy ? Since there are 2 vars, it's not possible to say that x=5 and y=2..


The Attempt at a Solution


D = ( 345 * 9 ) + (-9.8 * .5 * 9^2)


any help would be muc appreciated.
 
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Welcome to PF, Carbunkulous!
So close . . . just missed noticing that the distance down equals the distance up. That should eliminate one of the variables.
 
i've worked it down to

sqrt(345/4.9)+Y^2=81

does that make sense?, just solve that then 9-x to get the 2 times?
 
I don't see how you could have got that equation. And it doesn't give the right answer. Better go back to your
( 345 * 9 ) + (-9.8 * .5 * 9^2)
Put an equal sign in the middle instead of the + sign, and change the 9's since neither the up or the down time is 9. Rather, their total is 9.
 
Ok, me and my 2 friends have been working on it for about an hour now, and we've came down to the possible solutions:

303 meters
or
319 meters

can someone tell me which one is right, we have the math already figured out, just don't know which one is right. =/
 
Easy to check! Work with your 303 m first, find the time to fall using d=.5*a*t^2 solved for t, then add the time for the sound to come back up. If it doesn't work out to 9 s, then try the other answer!
You could actually solve it this way by trial and error - and it would be really fast if you had a spreadsheet set up to do all the calcs.
 
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