Question

Carol drops a stone into a mine shaft 184 meters deep. The average temperature of the air inside the shaft is 1.9 degrees Celsius. How soon after she drops the stone does she hear it hit the bottom of the shaft? Include units in your answer.

Answer 1

Given,

Distance, S = 184 m

The displacement equation is given by,

`S=ut+(1)/(2)gt^2`

Here, u is the initial velocity of the stone which is equal to zero, g is the acceleration due to gravity calculated as

`g=9.8m/s^2`

Substitute the given values,

`\begin{gathered} 184\text{ m = o+}(1)/(2)*9.8m/s^2* t^2 \\ t^2=(184*2)/(9.8) \\ t^2=37.55 \\ t=6.127\text{ s} \end{gathered}`

The fall time is 6.127 s.

Now,

`S=v* t`

Here, v is the speed of sound in the air 343 m/s.

`\begin{gathered} 184\text{ = 343}* t \\ t=(184)/(343) \\ t=0.536\text{ s} \end{gathered}`

Calculate the time from when she hears it hit the bottom of the shaft is

`\begin{gathered} T=6.127\text{ s + 0.536 s} \\ T=6.663\text{ s} \end{gathered}`