9514 1404 393
Answer:
4192.9 ft
Step-by-step explanation:
We assume your falling-distance formula is supposed to be ...
s = 16t²
So, the time required to fall distance s is ...
t = √(s/16) = (1/4)√s
The time required for sound to travel distance s is ...
s = 1100t
t = s/1100
Then the sum of the time for the ball to fall and the time for the sound to travel back is ...
(1/4)√s + s/1100 = 20
275√s = 22000 -s . . . . . multiply by 1100 and subtract s
75625s = s^2 -44000s +484,000,000 . . . square both sides
s^2 -119,625s +484,000,000 = 0 . . . . put in standard form
s ≈ (1/2)(119,625 ±√12,374,140,625) = {4192.943, 115432.057}
Only the smaller of these two solutions makes any sense in this problem.
The hole is about 4192.9 feet deep.
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<em>Additional comment</em>
The distance equation for the falling object presumes a vacuum. The sound transmission presumes the presence of air, so the question setup is self-contradictory.
