Question

A man drops a stone into a water well on his farm. He hears the sound of the splash 4.26s later. How deep is the well? The acceleration due to gravity is 9.8m/s^2 and the speed of sound in air is 313m/s. Answer in units of m

Answer 1

First let us calculate the time required for the stone to drop, say t1.

We use the formula:

h = vi t1 + 0.5 g t1^2

where h is height, vi is initial velocity = 0

h = 4.9 t1^2

 

Then the time required for the sound to go up, say t2:

h = 313 t2 – 4.9 t2^2

 

The two heights are equal so equate:

4.9 t1^2 = 313 t2 – 4.9 t2^2

 

We know that:

t1 + t2 = 4.26

so,

t1 = t2 – 4.26

 

Therefore:

4.9 (t2 – 4.26)^2 + 4.9 t2^2 – 313 t2 = 0

4.9 (t2^2 – 8.52 t2 + 18.1476) + 4.9 t2^2 – 313 t2 = 0

4.9 t2^2 – 41.748 t2 + 88.92324 + 4.9 t2^2 – 313 t2 = 0

9.8 t2^2 – 354.748 t2 = -88.92324

t2^2 – 36.2 t2 = -9.0738

 

Completing the square:

(t2 – 18.1)^2 = -9.0738 + 327.61

t2 – 18.1 = ± 17.85

t2 = 0.25s, 35.95

t2 cannot be greater than 4.26 s, therefore correct one is:

t2 = 0.25 s

 

Therefore height of the well is:

h = 313 t2 – 4.9 t2^2

h = 313 * 0.25 – 4.9 * 0.25^2

h = 77.94 m = 78m