Question

A stone is dropped into a well. The sound of the splash is heard 2.92 s later. What is the depth of the well? (Take the speed of sound to be 343 m/s.)

Answer 1

Answer :

Depth of well = 38.42 m

Let t₁ be the time taken for the stone to hit the well.

t₂ be the time taken for the sound to reach up to the ground level.

t₁ + t₂ = 2.92 s

So t₂ = 2.92 -t₁

Distance covered by stone to hit the well from ground level , s₁= ut +1/2 (at²)

Here initial velocity u= 0, a = acceleration due to gravity =g =9.8 m/s² , t= t₁

s₁ = 0+1/2(9.8t₁²) __________________(1)

Distance covered by sound wave to travel up to the ground level, s₂ = v*t

where v = speed of sound = 343 m/s

time ,t = t₂

So s2 = 343 t₂

Substituting t₂ = 2.92 -t_1,

s₂ = 343(2.92-t₁) __________________(2)

Since both the distances are equal ,s₁ = s₂

So, 1/2 (9.8 t1²) = 343(2.92 - t₁)   

4.9 t₁² = (343*2.92)- 343t₁   

4.9 t₁²+ 343 t₁ -1001.56 = 0

Solve this quadratic equation to get t_1.

To solve a quadratic equation of the form ax² +bx +c = 0

Here a =4.9 , b =343 , c= -1001.56

or

= 2.8 or -72.8

Negeative value for t1 can be omitted.

so t₁ =2.8 s

So depth of well is s₁ =1/2(9.8 t₁²)

=4.9 *2.8² = 38.42 m.