A man drops a rock into a well. (a) The man hears the sound of the splash 2.40s

Question

A man drops a rock into a well.
(a) The man hears the sound of the splash 2.40s after he releases the rock from rest. The speed of sound in air (at the ambient temperature) is 336 m/s. How far below the top of the well is the surface of the water?
(b) What if the travel time for the sound is ignored, what percentage error is introduced when the depth of the well is calculated? A man drops a rock into a well.
(a) The man hears the sound of the splash 2.40s after he releases the rock from rest. The speed of sound in air (at the ambient temperature) is 336 m/s. How far below the top of the well is the surface of the water?
(b) What if the travel time for the sound is ignored, what percentage error is introduced when the depth of the well is calculated? A man drops a rock into a well.
(a) The man hears the sound of the splash 2.40s after he releases the rock from rest. The speed of sound in air (at the ambient temperature) is 336 m/s. How far below the top of the well is the surface of the water?
(b) What if the travel time for the sound is ignored, what percentage error is introduced when the depth of the well is calculated?

Explanation / Answer

let the depth is h. then time taken to travel to the bottom=sqrt(2*h/g) time taken for sound to travel back=h/336 so sqrt(2*h/9.8)+(h/336)=2.7 solving for h, we get h=33.157 m b) if we ignore tavel time of sound,then sqrt(2*h/9.8)=2.7 h=35.721 m so %error=(35.721-33.157)*100/33.157=7.733 % please rate me...

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