Free-fall question with extras

[kylera]

This problem incorporates the main idea of gravitational freefall, but also adds an extra to it, and I get different answers from the back of the book:

A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard 3.2s later. If the speed of sound is 340m/s, how high is the cliff?

Book answer: 46m.
My answer: 52m.

My solution: Because of the lack of information, I simply used variables to substitute.
Given: x0 = 0, a = -9.8m/s^2, total time = 3.2s

x = x0 + v0t + .5at^2 = -4.9t^2 => I changed this to 4.9t^2 to reflect a change in point of reference.

Since the speed of sound is given, and x is assigned as above, I determined that the time in which sound traveled up the cliff is 340/x sec. Since the problem is asking for distance, I swapped the variables around with the length to get the following: t^2 = x/4.9 => (x/4.9)^.5 => time it takes to fall down the cliff.

Given that 3.2 s is the total time for the rock to fall AND for the sound to rebound, I wrote down the following: 3.2 = (340/x) + (x/4.9)^.5 . One calculation led to another, and I got the answer at top. Am I right this far and I got my variable switching wrong, or did I do something fundamentally wrong? Eagle eyes most apprecated :D

 

[Fermat]

The time for sound to travel x metres at 340 m/s is,

t = x[m]/340[ms^-1]
t = x/340
=========

 

[kylera]

The time for sound to travel x metres at 340 m/s is,
t = x[m]/340[ms^-1]
t = x/340
=========

Can you explain how you were able to determine that? I don't quite understand.

 

[civil_dude]

You are on the right track.

let the distance of the fall = d.

so d = 340(3.2-t) = 1/2(9.8)t^2

Use the quad. eq. to solve for t

ta da
t = 3.06s, so d = 45.93m

 

[kylera]

So it WAS my calculation! Damn, you made it look so simple! Thanks!