# Homework Help: How high is the cliff using speed of sound?

1. Sep 21, 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 ater, you 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 = VoT + 1/2aT^2
Vf = Vo + aT
Vf^2 = Vo^2 + 2a(deltaX)

V = (deltaX)/(deltaT) maybe?

3. The attempt at a solution

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?

am i close? where did i mess up?