A person drops a book into a well in a cave. He hears the sound of the book hitting the water 10 seconds later. The speed of the sound in air in the cave is 344m/s. How deep is the well?

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The person drops a book into a well in a cave. He hears the sound of the book hitting the water 10 seconds later. The speed of the sound in air in the cave is 344 m/s. Let the depth of the well be x m. If the time taken by the book to hit the water is t, `x = (1/2)*9.8*t^2`

=> `t = sqrt(x/4.9)`

The time taken by the sound to reach the person after it hits the water is `x/344` .

As the person hears the book hitting the water after 10 s,

`sqrt(x/4.9) + x/344 = 10`

=> `x/4.9 = (10 - x/344)^2`

=> `x/4.9 = 100 + x^2/118336 - (5x)/86`

The equation has two solutions `(16*1290^(3/2)+760240)/49` and `(-16*1290^(3/2)+760240)/49` . As the speed of sound is 344 m/s, the distance cannot be greater than 3440 m, this eliminates the root `(16*1290^(3/2)+760240)/49` . The approximate value of the other root is 386.16 m

**The depth of the well is 386.16 m**

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