a stone is dropped from rest into a well. the sound of the splash is heard exactly 2.00 s later. Find the dept

Question

a stone is dropped from rest into a well. the sound of the splash is heard exactly 2.00 s later. Find the depth of the well if the air temperature is 10.0 degrees Celsius.

Answer 1

The elapsed time is in two parts: the time for the stone to fall, and the time for the sound to get back to the top of the well. We know the sum of those times:

T1 + T2 = 2.00 sec.

To find T1, use kinematics:

d = voy T1 + 1/2 g T1^2

d is the depth of the well ( what you're looking for )
voy is the initial velocity in the vertical dimension ( zero )
g is the local acceleration of gravity ( 9.8 m/s^2 )

Solve for T1 to get

T1 = sqrt ( 2d / ( 9.8 )

Now for the return trip time T2. Sound travels through air at about 330 m/s, so

T2 = d / ( 330 )

Now substitute for T1 and T2 in the first equation and solve for d.

Answer 2

air temp will affect the speed of sound.

speed of sound is approx 338 m/s at 10 degrees C and seal level. that would reduce travel time by approx .03 seconds

about 19.03 m

equations are s=1/2 a t^2

where t is 2.0 - s/338

Answer 3

s= ut+1/2at^2
u is intial velocity = 0 because it was dropped
s = 0 + 1/2at^2
a = g =9.81m/s2
s = 4.9t^2
s = 4.9 x 4
= 19.6 m

air temp??? wont really affect the depth.
assumptions: negligible air resistance, dropped perfectly vertically, time taken for sound waves to travel back from source is negligible)