a stone is dropped In a well. the sound of the splash is heard after 2.5 s. find the depth of the well , take velocity of the sound =330m/s

As the sound heard after 2.5 s means,it has travelled distance equal to the height

The velocity of sound in air = 330 m/s

Time taken by the stone in reaching the well

Using the second equation of motion;

H=ut+1/2at^2

As initial vel=0,H=1/2gt^2

Time taken by the sound of splash to reach the top of the well,

t'=H÷330

=gt2÷(2×330)

=t2÷66

Given that,

t+t'=2.5

t+t2/66=2.5

Solving we get,t=2.41 s

and t'=0.09 s

.: H= 330×0.09

=29.04m//

Hope this answer helped.

Hi Ajay,

Here is the answer

Let total time from the point of dropping the stone and hearing the sound is T= t1+t2

Where

t1 time taken by stone to hit the water

t2 time taken by sound to reach the ground.

The distance travelled by sound = height of well =h = velocity of sound* time taken by sound to reach the ground

h= c*t2

h= 330t2 ……(1)

where c is velocity in m/s, t2 in sec. And h is in m.

Now dropping a stone is case of free fall under gravity.

Let initial velocity u=0

t1= time taken by stone to hit water

Hence

h=ut+1/2(gt^2)

h= 0+ 1/2 ( 10* (t1)^2)

Taking g= 10 m/s^2

h= 5(t1)^2……(2)

Equating 1 and 2

330 t2 = 5(t1^2)

But t1= T-t2= 2.5-t2

330t2 = 5((2.5-t2)^2)

330t2= 5( 6.25–5t2+ t2^2)

330t2=80–40t2+5t2^2

5t2^2–355t2+31.25=0

t2^2–71t2+6.25=0

We will get t2= 0.09 sec.

Hence h= 330*0.09= 29.04 m

Hope that answers your question