Question

Tinku dropped a stone into a very deep well. He hears the sound 6.53 s after he had dropped the stone. Find the depth of the well. Take acceleration due to gravity, g = 10m/s^{​​​​​​2} and Vsound as 340 m/s

Answer 1

The problem is in 2 parts:

1)The stone falls under gravity to the bottom of the well.

2)The sound travels back to the surface.

Let ,
t1=time taken to reach the bottom
t2=time taken for sound to return to tinku

We use the fact that the distance is common to both.

The distance the stone falls is given by:
(Towards well)
d=ut +(1/2)g(t^2)
[Initially at rest, so u=0]

ie, d=(1/2)g(t1^2).....(1)

We know that average speed = distance travelled / time taken.

We know the speed of sound =340 m/s .so we can say:
Distance traveled by sound towards tinku,d=velocity * time

d=340×t2........(2)

We know that:

t1+t2=6.53

We can put (1) equal to (2)⇒

∴340×t2=(1/2)g(t1^2)..... (3)

t2=(6.53−t1)

Substituting this into (3)⇒

340(6.53−t1)=(1/2)g(t1^2)

∴2220.2−340t1=(1/2)g(t1^2)

Let g=10m/s2

∴5(t1^2)+340t1−2220.2=0

This can be solved using the quadratic formula:

t1=
[−340±√{(340^2)−(4×5×2220.2)}]/(2*5)

Ignoring the -ve root this gives:

t1=6.0005xs

∴t2=6.53−6.0005=0.5295s

Substituting this back into (2)⇒

d=343×t2=343×0.5295=180.03m