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Chapter 2: Problem 49

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. Glenda drops a coin from ear level down a wishing well. The coin falls a distance of \(7.00 \mathrm{m}\) before it strikes the water. If the speed of sound is \(343 \mathrm{m} / \mathrm{s}\), how long after Glenda releases the coin will she hear a splash?

Short Answer

Expert verified
Answer: It takes Glenda 1.21 seconds to hear the splash after she drops the coin.

Step by step solution

01

Calculate the time it takes for the coin to fall

Since we know the distance the coin falls (\(7.00 \mathrm{m}\)) and the acceleration due to gravity (\(9.80 \mathrm{m} / \mathrm{s}^{2}\)), we can use the free-fall equation to find the time it takes for the coin to fall: $$h = \frac{1}{2}gt^2$$ We know h, and g, and we want to find t. To do this, rearrange the equation to solve for t: $$t = \sqrt{\frac{2h}{g}}$$ Now plug in the values for h and g: $$t = \sqrt{\frac{2(7.00 \mathrm{m})}{9.80 \mathrm{m} / \mathrm{s}^{2}}}$$
02

Calculate the time it takes for the sound to travel back to Glenda

To find the time it takes for the sound to travel back to Glenda, we can use the speed of sound (\(343 \mathrm{m} / \mathrm{s}\)). The time it takes for the sound to travel is given by the equation: $$t_s = \frac{d}{v}$$ where t_s is the time it takes for the sound to travel, d is the distance it travels, and v is the speed it travels. Plug in the values for d (same as h) and v: $$t_s = \frac{7.00 \mathrm{m}}{343 \mathrm{m} / \mathrm{s}}$$
03

Find the total time

To find the total time, add the time it takes for the coin to fall (t) and the time it takes for the sound to travel back (t_s): $$t_{total} = t + t_s$$ First, calculate the values for t and t_s obtained in the previous steps: $$t = \sqrt{\frac{2(7.00 \mathrm{m})}{9.80 \mathrm{m} / \mathrm{s}^{2}}} = 1.19 \mathrm{s}$$ $$t_s = \frac{7.00 \mathrm{m}}{343 \mathrm{m} / \mathrm{s}} = 0.020 \mathrm{s}$$ Now, add these values: $$t_{total} = 1.19 \mathrm{s} + 0.020 \mathrm{s} = 1.21 \mathrm{s}$$ Thus, Glenda will hear the splash 1.21 seconds after she releases the coin.

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