Question

A stone is dropped from rest into a well. The sound of the splash is heard exactly 3.90 s later. Find the depth of the well if the air temperature is 16.0°C.

Answer 1

The stone falling under gravity let t1 be the time from release to splash down

d = 1/2 g t²₁ ....1

after splash down the sound travels back up using 330 m/s for the speed of sound this gives

d = 330*t2 ......2

we also know that t1+t2 = 3.90....3

combine 1 and 2

330t2 = 1/2 g t²₁

from ..3

t2 = ( 3.9-t1)

so 330(3.90- t1) = 1/2 g t²₁

1287-330t1=1/2 g t²₁

_1/2 g t²_{1 + 330t1 -1287=0}

using quadratic equation formula to solve for t1

t1 = -330+- root 330^2-4*9.8/2*-1287 / 9.8

ignore the -ve root this gives t1 = 1.32

so t2 = 3.9-1.32 = 2.58

so d = 330*2.58 = 851.4 m