Trouble with exponents and logs

[corbell777]

Givens:
x = 29,300 ft
g = 32.2 ft/sec2
t = 116 sec
Solve for v:
x = (v2/g)ln(cosh(gt/v))
gx/v2 = ln(cosh(gt/v))
exp(gx/v2) = exp(gt/v)+exp(-gt/v)=exp(2gt/v+1)/2exp(gt/v)
2exp(gt/v)exp(gx/v2)=exp(2gt/v+ 1)
ln2 + gt/v + gx/v2=2gt/v + 1
v2ln2 + gtv + gx = 2gtv + v2
(ln2 - 1)v2 - gtv +gx = 0
v = {gt +/- sqrt[g2t2 - 4(ln2 - 1)gx]}/2(ln2 - 1)
v = {(32.2)*(116) +/- sqrt[(32.2)2*(116)2 - 4(-0.307)*(32.2)(29,300)]}/2(-0.307)
v = {3735.2 +/- sqrt[13951719 + 1158013]}/(-0.614)
v = (3735 +/- 3887)/(-0.614)
v = {-12414 ft/sec, 248 ft/sec}
Choose v positive gives an answer of v = 248 ft/sec.
However, the answer in the book is 265 ft/sec.
I've looked and looked and looked and I can't find my mistake.
I am not in a class, I am just studying on my own.
So any help would be much appreciated as I have no one else to ask!

P.S. I'm not sure how the box got around that one line.
I did not mean to emphasize it. It just happened.

 

[HallsofIvy]

I don't believe such a function, with v both inside cosh and outside it, can be solved in terms of elementary functions.

 

[mmm4444bot]

Weird. I responded in this thread yesterday, but today my post is gone. (Maybe I dreamt that I posted, lol.)

Anyway, I had substituted all the given values first, and I converted cosh to exponential form.

I got only one solution: -252.5862

Was the formula for x given to you, or is that equation something that you worked out yourself?

Halls is correct; the symbolic solution for v is written in terms of zeros of some families of functions (i.e., no closed-form solution is possible).

 

[corbell777]

This equation was given as the answer in the previous part of a problem in the book. Then we were to solve for v. Then saying terminal v = sqrt(mg/k) we were to find the drag coefficient k.