Sound: Stones falling from cliffs

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Homework Help Overview

The problem involves a stone dropped from a cliff, with the splash heard 2.1 seconds later. The context is kinematics and sound propagation, specifically calculating the height of the cliff based on the time taken for the stone to fall and the sound to travel back up.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equations used to relate the time of fall and the time for sound travel, questioning the sign convention and the validity of the results obtained. There are attempts to manipulate the equations to find a relationship between the variables.

Discussion Status

Some participants express confusion over the calculations and the results, particularly regarding the time taken for sound to travel. There is recognition that the derived time for sound travel seems inconsistent with the total time given. Participants are exploring the implications of their findings and questioning the assumptions made in their calculations.

Contextual Notes

There is an emphasis on the total time of 2.1 seconds being the sum of the time for the stone to fall and the time for the sound to return, leading to questions about the feasibility of the calculated times. The discussion reflects a need for clarity on the physical dimensions involved in the equations used.

Ritzycat
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Homework Statement


A stone is dropped from the top of a cliff. The splash it makes when striking the water below is heard 2.1s later. The speed of sound in air is 343 m/s. How high is the cliff?

Homework Equations


Δx=½at^2
t_1+t_2=2.1s
t_2=x/343m/s

The Attempt at a Solution


I used those equations to mathematically solve for the value of x. However I was getting bizarre answers when I solved the system of equations. t1 or t2 should not be more than 2.1s, as that is the total time from when it was dropped and the sound traveled back up. I set down as negative, so I used -9.8 for a.
 
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I'm sorry but i don't understand why you used a as -9.8m/s², if you took that as your sign convention,then your first equation would be,
-x=½(-9.8)t₁².
Post your calculation, you should be getting an equation with √x and x.
 
after some manipulation/substitution

343/t_2=0.5(9.8)(2.1-t_2)^2

t_2=5.6s

T2 being the amount of time it takes for the sound to travel back up to the top of the cliff. It should not be a longer amount of time than the time it takes for the stone to fall AND for the sound to come back up. (2.1s) I don't know what's wrong here!
 
Do you think 343/t2 has the dimension of a distance ? If sound travels 343 m/s, does it go 171.5 m in 2 seconds ?
 
Ritzycat said:
after some manipulation/substitution

343/t_2=0.5(9.8)(2.1-t_2)^2

t_2=5.6s

T2 being the amount of time it takes for the sound to travel back up to the top of the cliff. It should not be a longer amount of time than the time it takes for the stone to fall AND for the sound to come back up. (2.1s) I don't know what's wrong here!

This is a bit tricky. Can you estimate what h is?
 
Dear Ritz,

Physics is also about gut feeling. If something falls less than 2.1 seconds, it doesn't fall much further than ##{1\over 2}\; 9.81 \; 2^2 \approx 20## meter. So t2 is really small. If you get 5.6 seconds, you know it's wrong.

You knew it was wrong anyway, right ? I saw you write t1 + t2 = 2.1 somewhere and I don't think going back in time is in order here :wink:
 

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