Question

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 ?

Answer 1

`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"`

Answer 2

`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"`