How deep is the well

A man drops a stone into a water well on
his farm. He hears the sound of the splash
4.38 s later.
How deep is the well? The acceleration due
to gravity is 9.8 m/s2 and the speed of sound
in air is 333 m/s.
Answer in units of m

h = depth of water surface in the well (unknown)
u = speed of sound in air = 333 m/s
g = acceleration of gravity = 9.8 m/s^2
t1 = time of fall, down the well shaft (unknown)
t2 = sound travel time, up the well shaft (unknown)
t = total time = 4.38 s

h = (g/2)t1^2 . . . (1)
h = u t2 . . . . . . .(2)
t = t1 + t2 . . . . . (3)
..... 3 equations in 3 unknowns -- h, t1, t2

From (2): t2 = h/u
From (3): t = t1 + h/u; t1 = t - h/u
From (1): h = (g/2)(t - h/u)^2
... in which h is the only unknown; solve for it.

Multiply through by 2u^2 /g and rearrange:
h^2 - 2u(t + u/g)h + (ut)^2 = 0
h = u(t + u/g) ± √[(u(t + u/g))^2 - (ut)^2]
= u( (t + u/g) ± √[(t + u/g)^2 - t^2] )
In the limit where u→∞, the "+" choice will make h→∞, when it should go h→(g/2)t^2.
So only the "-" choice is valid:

h = u( (t + u/g) - √[(t + u/g)^2 - t^2] )
= 83.5 m

NB: the outer parentheses here contain t2 [evident from eqn. (2)], which is
t2 = 0.251 s, making the drop time
t1 = t - t2 = 4.13 s