Question

A stone is dropped into the well of depth 196m. The sound splash is heard after 6.88 sec.Find the velocity in the air????


Pls find and with explain

Answer 1

Time taken by the stone to reach the water surface  can be calculated using the equation of motion
s = ut + ½ gt2
Here u = 0, s = 196 m , g = 9.8 m/s2
 
So
t = (2x196/9.8)½
         = A
Therefore time taken by sound to travel this distance = 6.88 - A= B sec.
Speed of sound in air = distance covered/Time
                                            = 196 m/ (B s)
                                            = C m/s


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Answer 2

The velocity of sound in air is  350m/s.

The stone is dropped into a well of depth 196 m which means that the stone needs to cover a distance of 196 m in total to reach till the bottom of the well and for the splash sound to be heard.

The stone travels all the way down to the bottom of the well and this distance will be given by the depth of the well. Thus, we have:

$Distance=196\;m$

The splash sound however takes a certain period of time before reaching our ears, that is, due to echo there is a lag in the sound reaching us.

It is given that the splash sound which in this case is the echo sound is heard after 6.88 seconds.

In-order to calculate this lag we need to first calculate the time taken for the stone to reach the bottom of the well after which the sound is produced.

This time can be calculated by one of the equation of kinematics given below:

$s=ut+\frac{1}{2}\times gt^{2}$

where,

s is the distance

t is the time taken

g is the acceleration due to gravity

u  is the initial velocity

Here, the distance $s=196\;m$ and acceleration due to gravity is a constant and its value is  $g=9.8\;m/s^{2}$

The initial velocity of the stone is zero here because the stone is thrown freely or is falling freely so the body is initially said to be at rest.

Thus, we have: $u=0$

Substituting all these values in the above equation we get:

$196=(0)t+\frac{1}{2}\times (9.8)t^{2}$

$\implies 196=\frac{1}{2}\times (9.8)t^{2}$

$\implies 196=4.9t^{2}$

$\implies t^{2}=40\;s$

$\implies t=6.32\;s$

This time is the time taken by the stone to cover the distance 196 m to reach the bottom of the well. It is given that the splash sound is 6.88 s. Thus, the time taken by the sound to reach the us will be:

$t=6.88-6.32$

$\implies t=0.56\;s$

Hence, the velocity of the sound waves is given by the equation:

$Velocity=\frac{Distance}{Time}$

By substituting the distance and the time values we get:

$Velocity=\frac{196}{0.56}$

$\implies Velocity=350\;m/s$

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