Timed freefall + sound to respond from cave

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SUMMARY

The discussion centers on a physics problem involving a rock dropped into a deep abyss, where sound travels back to the observer at a speed of 345 m/s. The total time measured for the rock to fall and the sound to return is 9.00 seconds. Participants concluded that the depth of the abyss can be calculated by determining the time taken for the rock to fall (x) and the time for the sound to travel back (9.00 - x). The potential solutions for the depth were identified as either 303 meters or 319 meters, with a recommendation to verify the correct answer by calculating the time for each depth option.

PREREQUISITES
  • Understanding of kinematic equations, specifically d = 0.5 * a * t^2
  • Knowledge of the speed of sound in air (345 m/s)
  • Basic algebra for solving equations with two variables
  • Familiarity with the concept of free fall and gravitational acceleration (9.8 m/s²)
NEXT STEPS
  • Calculate the time of fall using the equation d = 0.5 * a * t^2 for both 303 m and 319 m
  • Learn about the relationship between distance, speed, and time in physics
  • Explore the use of spreadsheets for solving physics problems through trial and error
  • Study the principles of sound propagation and its implications in physics problems
USEFUL FOR

High school physics students, educators teaching kinematics, and anyone interested in solving real-world physics problems involving free fall and sound travel.

carbunkulous
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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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