Sound: Stones falling from cliffs

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SUMMARY

The problem involves calculating the height of a cliff from which a stone is dropped, with the splash heard 2.1 seconds later. The speed of sound in air is given as 343 m/s, and the equations used include Δx=½at² and t₁+t₂=2.1s. The calculations reveal that the time for sound to travel back (t₂) cannot exceed the total time of 2.1 seconds, indicating an error in the initial assumptions. The correct approach involves estimating the height using the equation h = ½gt², leading to a maximum height of approximately 20 meters.

PREREQUISITES
  • Understanding of kinematic equations, specifically Δx=½at²
  • Knowledge of the speed of sound in air (343 m/s)
  • Basic principles of physics related to free fall and sound propagation
  • Ability to manipulate and solve algebraic equations
NEXT STEPS
  • Review kinematic equations for free-fall motion
  • Study the relationship between time, distance, and speed in sound propagation
  • Explore the concept of sign conventions in physics problems
  • Practice solving similar physics problems involving sound and motion
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Students studying physics, educators teaching kinematics, and anyone interested in solving real-world problems involving motion and sound.

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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