Question

A stone is dropped from rest into a well. The sound of the splash is heard exactly 2.30 s later. Find the depth of the well if the air temperature is 6.0°C. m

Answer 1

speed of sound v = 331 + (0.6*T)

v = 331 + (0.6*6)

v = 334.6 m/s

for the stone

displacement y = -h

initial velocity voy = 0

acceleration ay = -g

y = voy*t + (1/2)*ay*t1^2

-h = -(1/2)*g*t1^2

h = (1/2)*g*t1^2

h= (1/2)*9.8*t1^2

h = 4.9*t1^2 .......( 1 )

for the sound

distance travelled h = v*t2

h = 334.6*t2 ............( 2 )

given

t1 + t2 = 2.3 s

from 1 and 2

4.9*t1^2 = 334.6*t2

4.9*t1^2 = 334.6*(2.3-t1)

t1 = 2.23 s

t2 = 2.3 - 2.23 = 0.07 s

depth d = 334.6*0.07 = 23.422 m