How high is the cliff and speed of sound?

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Homework Help Overview

The problem involves determining the height of a cliff by dropping a rock and measuring the time until the sound of the impact is heard. The scenario considers the effects of gravity and the speed of sound, with a total time of 10 seconds for both the rock's fall and the sound's travel back up.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between the time it takes for the rock to fall and the time for the sound to travel back up. There are attempts to set up equations for both the rock's motion and the sound's travel, leading to questions about how to account for the total time and the unknown variables involved.

Discussion Status

There is ongoing exploration of the equations needed to relate the height of the cliff to the time variables. Some participants express confidence in their progress while also acknowledging uncertainties about the relationships between the variables. Guidance has been offered regarding the setup of equations and the substitution of variables.

Contextual Notes

Participants note the challenge of having two unknowns (the time for the rock to fall and the time for the sound to travel) and the need to express one in terms of the other. The discussion reflects a collaborative effort to clarify the problem setup without arriving at a definitive solution.

katamoria
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Homework Statement


To find the height of a cliff, you drop a rock from the top, and 10s later, hear the sound of it hitting the ground at the foot of the cliff.
ignoring air resistance, how high is the cliff if the speed of sound is 330m/s



Homework Equations



Xf = Xo +Vo(t) + 1/2at^2
Vf = Vo + at
Vf^2 = Vo^2 + 2a(Xf - Xo)

The Attempt at a Solution


I don't know how to account for the time it takes sound to travel. If I solve for Vf or Xf using these equations, then I get a number I know is for the ball being lower than the foot of the cliff. When I tried, using speed of sound, I don't know how long the sound is travelling, or how far it has to go.
 
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Do you have an equation relating the speed of sound, the height of the cliff, and how long it takes the sound to travel the height of the cliff?
 
Since it is all in one direction, straight path, the speed of sound is the velocity. If I plug that in, I still have two variables, time it takes for sound to travel back up the cliff, and how high the cliff is.

it is 10s total for ball to hit bottom, and sound to come back up, I can't just plug in 10s.
 
katamoria said:
Since it is all in one direction, straight path, the speed of sound is the velocity. If I plug that in, I still have two variables, time it takes for sound to travel back up the cliff, and how high the cliff is.

it is 10s total for ball to hit bottom, and sound to come back up, I can't just plug in 10s.

You're correct that there are two variables, the ones you said.

However, there are two equations you can write. One is for the rock dropping from the top of the cliff with an acceleration of -g. The other equation is for the sound traveling up the cliff at a constant speed of 330m/s. Writing out those two equations, and working with them, is how to solve this problem.
 
ok so here is what i tried.
i drew it so that
T1 = time for rock to fall
T2 = time for sound to go back
Tt = Total time = 10s
so
T1+T2 = Tt = 10s

then I did
Vs = velocity of sound, so
Vs = deltaX / deltaT
solved for deltaX which = height of the cliff and got

deltaX = Vs(Tt-T1)

then for the rock,
Vf^2 = Vo^2 +2a(deltaX)
knowing that Vo^2 = 0, and a = g, and solving for deltaX
deltaX = Vf^2/2g

since both equations are solving for the same deltaX, I set them equal to each other
Vs(Tt-T1) = Vf^2/2g

the unknowns are T1, and Vf

I feel pretty good up to this point, then I'm not so sure, but next i tried...
to solve for T2 by
Vf = Vo +at
t is for the rock so...
Vf = g(Tt-T2)

then plug that in and get the equation

Vs(Tt-T1) = g(Tt-T2)^2/2g
but there's still two unknowns with T1 and T2, so...
if (Tt-T2) = T1, can i plug that in on the right side, so that T1 is the only variable? it's just on both sides?
 
I really don't think i did that right...
 
You have made a lot of progress and are nearly there (I think) :smile:

katamoria said:
ok so here is what i tried.
i drew it so that
T1 = time for rock to fall
T2 = time for sound to go back
Tt = Total time = 10s
so
T1+T2 = Tt = 10s

then I did
Vs = velocity of sound, so
Vs = deltaX / deltaT
solved for deltaX which = height of the cliff and got

deltaX = Vs(Tt-T1)

then for the rock,
Vf^2 = Vo^2 +2a(deltaX)
knowing that Vo^2 = 0, and a = g, and solving for deltaX
deltaX = Vf^2/2g

since both equations are solving for the same deltaX, I set them equal to each other
Vs(Tt-T1) = Vf^2/2g

the unknowns are T1, and Vf

I feel pretty good up to this point, then I'm not so sure, but next i tried...
to solve for T2 by
Vf = Vo +at
t is for the rock so...
Vf = g(Tt-T2)
Okay, or you could also say

Vf = g T1​

then plug that in and get the equation

Vs(Tt-T1) = g(Tt-T2)^2/2g
Small error: the right hand side here should be

g^2 (Tt-T2)^2 / 2g​

since you are really substituting

Vf^2 → g^2 (Tt-T2)^2​

but there's still two unknowns with T1 and T2, so...
if (Tt-T2) = T1, can i plug that in on the right side, so that T1 is the only variable? it's just on both sides?

Yes, you can definitely do that ... and you should get the same thing as if you had used

Vf = g T1​

earlier.

Looks like you are nearly done, just need to solve for T1.
 

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