A stone is dropped into a well of depth 11 m. How long after that, the sound of the splash can be heard? Given speed of sound =343ms^-1

To find out how long it takes for the sound of the splash to be heard after the stone is dropped into the well, we can use the formula:

Time = Distance / Speed

In this case, the Distance is the depth of the well, which is 11 m, and the Speed is the speed of sound, which is given as 343 m/s.

Plugging in the values into the formula:

Time = 11 m / 343 m/s

Now we can calculate the time it takes for the sound of the splash to be heard:

Time = 0.0321 s

Therefore, the sound of the splash can be heard approximately 0.0321 seconds after the stone is dropped into the well.

To find out how long it takes for the sound of the splash to be heard, we need to calculate the time taken for the stone to fall into the well and then back up to reach the surface where the person is located.

Let's assume the stone falls with uniform acceleration due to gravity (9.8 m/s^2).

The first step is to find the time it takes for the stone to hit the bottom of the well. We can use the equation of motion:

s = ut + (1/2)at^2

Where:
- s is the distance (depth of the well) = 11m
- u is the initial velocity (0 since the stone is dropped)
- a is the acceleration due to gravity = 9.8m/s^2
- t is the time taken

Plugging the values into the equation, we get:

11 = 0*t + (1/2)(9.8)t^2

Simplifying the equation:

11 = (4.9)t^2

t^2 = 11/4.9

t^2 ≈ 2.24

t ≈ √2.24

t ≈ 1.50 seconds (approx.)

So, it takes about 1.50 seconds for the stone to hit the bottom of the well.

Now, the sound needs to travel from the bottom of the well back up to reach the person. The speed of sound is given as 343m/s.

The distance the sound needs to travel is also 11m (the depth of the well).

Using the formula: time = distance / speed

t = 11m / 343m/s

t ≈ 0.032 seconds (approx.)

Therefore, the sound of the splash can be heard approximately 0.032 seconds after the stone hits the bottom of the well.

how long does it take to fall 11m?

1/2 at^2 = 11

Then it takes the sounds 11/343 s to come back.