Height of cliff using energy and speed of sound

[toothpaste666]

Homework Statement

[/B]
A stone is dropped from the top of a cliff. The splash it makes when striking the water below is heard 4.0s later.
how high is the cliff

Homework Equations

.5mv^2+mgy = E
v = v0 + at
v = d/t

The Attempt at a Solution

The question says after the rock is DROPPED so i need to figure out the time it took to drop and the time it took for the sound to travel back and those two times added together will be 4 sec
t1 is the time it takes for the rock to hit the water. using energy
mgh = .5mv^2
gh = .5v^2
to find v
v = v0 + at
v = gt
so
gh = .5g^2t^2
2h=gt^2
2h/g = t^2
t1= sqrt (2h/g)

to find t2
v = d/t
v=h/t
t2= h/v where v is speed of sound

t1 + t2 = 4
sqrt(2h/g) + h/v = 4
sqrt(2h/g) = 4 - h/v
2h/g = (4-h/v)^2
2h/g = 16 - 8h/v + h^2/v^2
2h/g +8h/v = 16 - h^2/v^2
2hv/gv + 8hg/gv = 16 - h^2/v^2
[(2hv+8hg)/gv] = 16 - h^2/v^2
[(2v+8g)/gv]h = 16 - h^2/v^2
h^2/v^2 + [(2v+8g)/gv]h - 16 = 0
(1/v^2)h^2 +[(2v+8g)/gv]h -16 = 0
(1/(343^2))h^2 + [(2*343+8*9.8)/(9.8*343)]h -16 = 0
(8.50x10^-6)h^2 + .227 h -16 = 0
using quadratic formula
[-.227 +/- sqrt((-.227^2)-4(8.50x10^-6)(-16))]/(2* (8.50x10^-6)
(-.227 +/- .228)/ (1.7*10^-5)

-.227 + .228/ (1.7*10^-5) = 58.8m = h

mastering physics wants it to 2 sig figs so i submitted 59 m but it was marked wrong. where did i make a mistake?

 

[haruspex]

Have you tried substituting your answer back into the starting equations to calculate the two times?

 

[toothpaste666]

t1 = sqrt(2h/g) = sqrt((2*58.8)/9.8) = 3.46
t2 = h/v = 58.8/343 = .171
t1 + t2 = 4
3.46 + .171 = 3.63
so its not quite 4

 

[haruspex]

t1 = sqrt(2h/g) = sqrt((2*58.8)/9.8) = 3.46
t2 = h/v = 58.8/343 = .171
t1 + t2 = 4
3.46 + .171 = 3.63
so its not quite 4

Well short of 4. So try subbing it in the last quadratic equation. Narrow down the location of the error that way.

 

[toothpaste666]

ended up finding it by trial and error that way. still trying to figure out where i went wrong mathematically