First let us calculate the time required for the stone to drop, say t1.
We use the formula:
h = vi t1 + 0.5 g t1^2
where h is height, vi is initial velocity = 0
h = 4.9 t1^2
Then the time required for the sound to go up, say t2:
h = 313 t2 – 4.9 t2^2
The two heights are equal so equate:
4.9 t1^2 = 313 t2 – 4.9 t2^2
We know that:
t1 + t2 = 4.26
so,
t1 = t2 – 4.26
Therefore:
4.9 (t2 – 4.26)^2 + 4.9 t2^2 – 313 t2 = 0
4.9 (t2^2 – 8.52 t2 + 18.1476) + 4.9 t2^2 – 313 t2 = 0
4.9 t2^2 – 41.748 t2 + 88.92324 + 4.9 t2^2 – 313 t2 = 0
9.8 t2^2 – 354.748 t2 = -88.92324
t2^2 – 36.2 t2 = -9.0738
Completing the square:
(t2 – 18.1)^2 = -9.0738 + 327.61
t2 – 18.1 = ± 17.85
t2 = 0.25s, 35.95
t2 cannot be greater than 4.26 s, therefore correct one is:
t2 = 0.25 s
Therefore height of the well is:
h = 313 t2 – 4.9 t2^2
h = 313 * 0.25 – 4.9 * 0.25^2
h = 77.94 m = 78m