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Chapter 2: Problem 55

please assume the free-fall acceleration \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) unless a more precise value is given in the problem statement. Ignore air resistance. You drop a stone into a deep well and hear it hit the bottom 3.20 s later. This is the time it takes for the stone to fall to the bottom of the well, plus the time it takes for the sound of the stone hitting the bottom to reach you. Sound travels about \(343 \mathrm{m} / \mathrm{s}\) in air. How deep is the well?

Short Answer

Expert verified
Answer: The approximate depth of the well is 26.5 meters.

Step by step solution

01

Determine the time it takes for the sound to travel

We know the total time is 3.20 s and the speed of sound is 343 m/s. Let the time the sound takes to travel be \(t_{sound}\). Let the depth of the well be \(d\). We have the equation: $$d = 343 * t_{sound}$$ We will solve for \(t_{sound}\) after finding the total time for the stone to fall, which is the difference between the total time and the time of the sound.
02

Determine the time it takes for the stone to fall

Let the time it takes for the stone to fall be \(t_{fall}\). We can now find \(t_{fall}\) using the following expression: $$t_{fall} = 3.20 - t_{sound}$$
03

Write the equation for the distance of a free-falling object

We can use the free-fall equation to relate the depth of the well \(d\) to \(t_{fall}\). The equation for the distance of a free-falling object is given by: $$d = \dfrac{1}{2}gt_{fall}^2$$ Where \(g = 9.80 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity.
04

Substitute the expressions of \(t_{sound}\) and \(t_{fall}\) and solve for \(d\)

We have two equations: 1) $$d = 343 * t_{sound}$$ 2) $$d = \dfrac{1}{2}g(3.20 - t_{sound})^2$$ Substitute expression 1) into expression 2): $$343 * t_{sound} = \dfrac{1}{2}g(3.20 - t_{sound})^2$$ Now we have an equation with only one variable, \(t_{sound}\).
05

Solve the equation for \(t_{sound}\)

Solve the equation for \(t_{sound}\): $$343 * t_{sound} = \dfrac{1}{2}*9.80*(3.20 - t_{sound})^2$$ Solving for \(t_{sound}\), we get: $$t_{sound} \approx 0.883 \mathrm{s}$$
06

Calculate the time it takes for the stone to fall and the depth of the well

Now that we have \(t_{sound}\), we can calculate \(t_{fall}\): $$t_{fall} = 3.20 - t_{sound} \approx 3.20 - 0.883 = 2.317 \mathrm{s}$$ Using the time \(t_{fall}\), we can find the depth of the well \(d\) using the free-fall equation: $$d = \dfrac{1}{2}g*t_{fall}^2 = \dfrac{1}{2}*9.80*(2.317)^2 \approx 26.5 \mathrm{m}$$ So, the depth of the well is approximately 26.5 meters.

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Most popular questions from this chapter

In the problems, please assume the free-fall acceleration $g=9.80 \mathrm{m} / \mathrm{s}^{2}$ unless a more precise value is given in the problem statement. Ignore air resistance. A car traveling at \(29 \mathrm{m} / \mathrm{s}(65 \mathrm{mi} / \mathrm{h})\) runs into a bridge abutment after the driver falls asleep at the wheel. (a) If the driver is wearing a seat belt and comes to rest within a 1.0 -m distance, what is his acceleration (assumed constant)? (b) A passenger who isn't wearing a seat belt is thrown into the windshield and comes to a stop in a distance of \(10.0 \mathrm{cm} .\) What is the acceleration of the passenger?
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