# You drop a stone down a well. You hear the splash 2.03 s later. How deep is the well? The speed of sound in air is 343 m/s.

*print*Print*list*Cite

A stone is dropped down a well and the sound of the splash when it strikes the water is heard 2.03 s later. The depth of the well has to be determined.

Let the depth of the well be L meters. The acceleration of the stone due to the gravitational force of attraction is assumed to be constant at 9.8 m/s^2. The speed of sound in air is 343 m/s.

As the stone falls it takes a time t to strike the water below. The equation `L = 0*t + (1/2)*(9.8)*t^2` shows the relation between the depth of the well and the time taken for the stone to fall.

After the stone strikes the water, the time taken by sound to travel upwards and be heard is `L/343` s. This gives us two equations.

`L/343 + t = 2.03` ...(1)

`L = 0*t + (1/2)*(9.8)*t^2` ...(2)

From (1)

`L = (2.03 - t)*343`

Substituting this in (2)

`(2.03 - t)*343 = (1/2)*(9.8)*t^2`

`2.03*343 - 343*t = 4.9*t^2`

Solving this equation gives `t = +-(21*sqrt(310)-350)/10` . The negative root can be ignored.

Substituting `t = +-(21*sqrt(310)-350)/10` in `L = (2.03 - t)*343` gives:

`L = (2.03 -(21*sqrt(310)-350)/10)*343`

L = 19.0998 m

**The depth of the well is 19.0998 m.**