Question

the Splash is heard after 2.05 seconds after the stone is dropped into a well of depth 19.6 metre the velocity of sound is:

Answer 1

HOPE IT MAY HELP YOU.





PLEASE MARK IT AS BRAINLIST

Answer 2

Answer:

The velocity of sound is $392\ m/s^{2}$

Explanation:

We need to calculate the time taken by the stone to reach the well at first and deduct that amount from the total time, which is 2.05s. Depth of the well is given, therefore, this remaining time was taken by sound to reach the observer, and hence we can find the

In the question, it is given that the splash is heard after 2.05s, this means that the stone reached the well and then the sound travelled back to the observer, this took 2.05s

The following values are given to us:

Time after which the splash is heard =2.05s(t)

Depth of the well =19.6m

$g =9.8 ms^{-2}$

Let t1  be the time taken by the stone to reach the water.

And h  be the depth of the well =19.6m

We know, the three equations of motion, from there let us consider the equation which helps us to calculate the distance traveled from initial velocity, acceleration, and time taken.

$h=ut+\frac{1}{2} gt_{1}^{2}$

Since the stone is thrown, we consider the initial velocity to be zero.

∴$U=0$

So, $h=\frac{1}{2} gt_{1}^{2}$

Now, rearranging the equation, we get:

$t_{1} =\sqrt\frac{2h}{g}$

Putting the values, we obtain:

$t_{1} =\sqrt\frac{2*19.6}{9.8}$

On solving, we get:

$t_{1} =4s$

Total time taken by the splash to reach the observer is =2.05s

, therefore, time taken for the sound to travel to the observer is: $t_{2} =t-t_{1}$

Therefore, $t_{2} =(2.05-2)s=.5s$

We know. Distance travelled by the sound is depth of the well, which is $=19.6m$

Thus, we can calculate the speed of sound, as we know:

⇒ $Speed=\frac{Distance travelled}{Time Taken}$

Putting the values, we obtain:

⇒ $Speed=\frac{19.6}{0.5}$

Thus, we finally arrive at:

⇒ $Speed=392m/s$

This is the required solution.