You drop a stone into a deep well and hear it hit the bottom 3.20 seconds later. This is the time it takes for the stone to fall to the bottom of the well, plus the time it takes for the sound to reach you. If sound travels at a rate of 343m/s in (cont.)?

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air, how deep is the well?

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Answer 1 · 2017-02-13T18:15:34

46.3 m

The problem is in 2 parts:

We use the fact that the distance is common to both.

The distance the stone falls is given by:

#sf(d=1/2"g"t_1^2" "color(red)((1))#

We know that average speed = distance travelled / time taken.

We are given the speed of sound so we can say:

#sf(d=343xxt_2" "color(red)((2)))#

We know that:

#sf(t_1+t_2=3.2s)#

We can put #sf(color(red)((1)))# equal to #sf(color(red)((2))rArr)#

#:.##sf(343xxt_2=1/2"g"t_1^2" "color(red)((3)))#

#sf(t_2=(3.2-t_1))#

Substituting this into #sf(color(red)((3))rArr)#

#sf(343(3.2-t_1)=1/2"g"t_1^2)#

#:.##sf(1097.6-343t_1=1/2"g"t_1^2)#

Let #sf("g"=9.8color(white)(x)"m/s"^2)#

#:.##sf(4.9t_1^2+343t_1-1097.6=0)#

This can be solved using the quadratic formula:

#sf(t_1=(-343+-sqrt(117,649-(4xx4.9xx-1097.6)))/(9.8)#

Ignoring the -ve root this gives:

#sf(t_1=3.065color(white)(x)s)#

#:.##sf(t_2=3.2-3.065=0.135color(white)(x)s)#

Substituting this back into #sf(color(red)((2))rArr)#

#sf(d=343xxt_2=343xx0.135=46.3color(white)(x)m)#