A stone is dropped from rest into a well. The sound of the splash is heard exactly 2.00s later. Find the depth of the well if the air temperature is 10.0°c ?

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Michael Share
Aug 27, 2015

Answer:

#19.8"m"#

Explanation:

For the 1st part the stone is falling under gravity. Let #t_1# be the time from release to splashdown.

#d# =depth

So #d=(1)/(2)"g"t_(1)^2# #" "##color(red)((1))#

After splashdown the sound travels back up. Using #330"m/s"# for the speed of sound this gives:

#d=330xxt_2##" "# #color(red)((2))#

We also know that:

#t_1+t_2=2"s"# #" "color(red)((3))#

Combining #color(red)((1))"and"color(red)((2))# we get:

#330t_2=1/2g.t_1^2#

From #color(red)((3))rArr#

#t_2=(2-t_1)#

So:

#330(2-t_1)=1/2."g"t_1^2#

#660-330t_1=1/2."g"t_1^2#

#1/2"g"t_1^2+330t_1-660=0#

Using the quadratic formula to solve for #t_1#:

#t_1=(-330+-sqrt((330)^2-4xx9.8/2xx(-660)))/(9.8)#

Ignoring the -ve root this gives:

#t_1=1.94"s"#

So #t_2=2-1.94=0.06"s"#

So #d=330xxt_2=330xx0.06=19.8"m"#

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Write your answer here...
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Answer

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Alan P. Share
Aug 26, 2015

Answer:

#674# meters

Explanation:

Although there will be some variation depending upon air humidity (for example) the speed of sound relative to temperature (measured in degrees Celsius) can be determined by the empirical formula:
#color(white)("XXX")s=331" meters"/"sec."+0.6 "meters"/"sec."*c#
where #c# is the temperature in degrees Celsius.

For a speed of #s# and time of #t#,
the distance #d# is
#color(white)(*XXX")d = sxxt#

Therefore we have
#color(white)("XXX")D =331+0.6(10) "meters"/"second" xx 2 "seconds"#

#color(white)("XXXX")= 674 "meters"#

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