At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Connect with a community of experts ready to help you find accurate solutions to your questions quickly and efficiently. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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 :

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.

Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.