# How high is the cliff and speed of sound?

1. Sep 19, 2009

### katamoria

1. The problem statement, all variables and given/known data
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

2. Relevant equations

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

3. 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.

Last edited: Sep 19, 2009
2. Sep 19, 2009

### Redbelly98

Staff Emeritus
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?

3. Sep 19, 2009

### katamoria

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.

4. Sep 19, 2009

### Redbelly98

Staff Emeritus
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.

5. Sep 21, 2009

### katamoria

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 eachother
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?

6. Sep 21, 2009

### katamoria

I really don't think i did that right...

7. Sep 21, 2009

### Redbelly98

Staff Emeritus
You have made a lot of progress and are nearly there (I think)

Okay, or you could also say

Vf = g T1​

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​

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.