Math Problem Statement

One method for determining the depth of a well is to drop a stone into it and then measure the time it takes until the splash is heard. If d is the depth of the well (in feet) and t1 the time (in seconds) it takes for the stone to fall, then d = 16t12, so t1 =

d /4. Now if t2 is the time it takes for the sound to travel back up, then d = 1090t2 because the speed of sound is 1090 ft/s. So t2 = d/1090. Thus, the total time elapsed between dropping the stone and hearing the splash is t1 + t2 =

d 4 + d 1090 . How deep is the well if this total time is 5 s? (See the following figure. Round your answer to one decimal place.)

Solution

We are tasked with determining the depth dd of a well, given that the total time between dropping a stone and hearing the splash is 5 seconds. The total time is the sum of two components:

  • t1t_1 is the time it takes for the stone to fall down the well.
  • t2t_2 is the time it takes for the sound to travel back up the well.

We are provided with the following relationships:

  1. The time for the stone to fall is t1=d4t_1 = \frac{d}{4}.
  2. The time for the sound to travel back is t2=d1090t_2 = \frac{d}{1090}.

The total time elapsed is the sum of these two times: t1+t2=d4+d1090t_1 + t_2 = \frac{d}{4} + \frac{d}{1090} We are also given that the total time is 5 seconds, so: d4+d1090=5\frac{d}{4} + \frac{d}{1090} = 5

Step 1: Combine the terms with a common denominator

To combine the fractions, we find a common denominator. The least common denominator between 4 and 1090 is 4360. Rewriting both fractions:

d4=1090d4360,d1090=4d4360\frac{d}{4} = \frac{1090d}{4360}, \quad \frac{d}{1090} = \frac{4d}{4360}

Now we can combine the fractions: 1090d+4d4360=5\frac{1090d + 4d}{4360} = 5 1094d4360=5\frac{1094d}{4360} = 5

Step 2: Solve for dd

Now, multiply both sides of the equation by 4360 to eliminate the denominator: 1094d=5×43601094d = 5 \times 4360 1094d=218001094d = 21800

Now, divide both sides by 1094: d=21800109419.92d = \frac{21800}{1094} \approx 19.92

Final Answer:

The depth of the well is approximately d=19.9d = 19.9 feet.

Would you like more details or clarifications on any part of the solution?

Here are 5 related questions to further your understanding:

  1. What assumptions are made in this problem about the effect of air resistance?
  2. How would the solution change if the speed of sound was different, such as in warmer or colder air?
  3. What formula would apply if the well were on a different planet with a different gravitational constant?
  4. How would you solve this if the total time was not 5 seconds, but another value?
  5. How does the speed of sound in water compare to that in air, and how would this affect the calculation if the splash occurred underwater?

Tip: Always pay attention to units, especially when mixing speeds (like sound) and distances (like depth), as they must be consistent to avoid errors!

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Math Problem Analysis

Mathematical Concepts

Algebra
Fraction Addition
Physics (Kinematics)
Units Conversion

Formulas

d = 16t1^2
t1 = d/4
t2 = d/1090
t1 + t2 = d/4 + d/1090

Theorems

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Suitable Grade Level

Grades 10-12