Question 515221: You drop a rock into a deep well. You can't see the rock's impact at the bottom, but you hear it after 4 seconds.The time that passes after you drop the rock has two components: the time it takes the rock to reach the bottom of the well, and the time that it takes the sound of the impact to travel back to you. Assume the speed of sound is 1100 feet per second. d=16T^2
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! You drop a rock into a deep well. You can't see the rock's impact at the bottom, but you hear it after 4 seconds.The time that passes after you drop the rock has two components: the time it takes the rock to reach the bottom of the well, and the time that it takes the sound of the impact to travel back to you. Assume the speed of sound is 1100 feet per second. d=16T^2
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On the way down distance = 16t^2
On the way up distance = 1100(4-t)
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Equation:
distance down = distance up
16t^2 = 4400 - 1100t
16t^2 + 1100t - 4400 = 0
4t^2 + 275t - 1100 = 0
t = 3.791 sec.
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distance down = 16*3.791^2 ~ 230 ft.
distance up = 1100(4-3.791) = 1100(0.2090) ~ 230 ft.
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Cheers,
Stan H.
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