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A rubber ball is dropped from the top of a hole. Exactly 20 seconds later, the sound of the rubber ball hitting bottom is heard. How deep is the hole? (Hint The distance that a dropped object falls in t seconds is represented by the formula s= 167. The speed of sound is 1100 ft/sec.)​

Sagot :

sqdancefan

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.

_____

Additional comment

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.

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