Depth of a Water Well by dropping a rock

[MienTommy]

Homework Statement

A man drops a rock into a well. He hears the sound of the splash 3.2 seconds after he releases the rock from rest. The speed of sound in air (at the local ambient condition) is 338 m/s

(a) How far below the top of the well is the surface of the water? (round your answer to a whole number)

(b) If you ignored the travel time for the sound, what would have been the calculated depth? (round your answer to a whole number)

time of rock + time of sound = 3.2s
acceleration = -9.8 m/s^2
V_i = 0 m/s
V_sound = 338 m/s

Homework Equations

x=x_0 + v_0*t + (1/2)at^2

The Attempt at a Solution

x = 0 + 0 + (1/2)(-9.8)(3.2)^2

However, I realized, this is incorrect because I also need to account the time the sound took to travel to the person's ear subtracted by the 3.2 seconds. My problem is, how do I find the time it took to reach the person's ear when the rock hit the surface of the water?

 

[voko]

Obviously, the rock takes time $t_1$ to hit the water, and the sound takes time $t_2$ to reach you. As you wrote, $t_1 + t_2 = t = 3.2 \ \text{s}$.

 

[PeroK]

You want to find the depth of the well, d. So, try to derive an equation that relates t and d.

For example, if d = 10m, it would be quite easy to work out t. So, try to use this approach to relate d and t generally.

 

[Abaidullah]

Hi
There r also some different issues like drag force of water on shape of rock so how this is possible

 

[HalfLostAndSearching]

I would approach this problem by first identifying the known and unknown variables and then using the appropriate equations to solve for the unknowns. In this case, the known variables are the speed of sound in air, the time it took for the sound to reach the person's ear, and the acceleration due to gravity. The unknown variable is the depth of the water well.

To find the depth of the water well, we can use the equation x=x_0 + v_0*t + (1/2)at^2, where x is the final position, x_0 is the initial position (in this case, the top of the well), v_0 is the initial velocity (which is 0 m/s for the rock), t is the time it took for the rock to hit the water, and a is the acceleration due to gravity.

(a) Using the given information, we can solve for the depth of the water well by rearranging the equation to x = (1/2)at^2 and plugging in the values. We get x = (1/2)(-9.8)(3.2)^2 = 50.24 m. However, since we are rounding to a whole number, the depth of the water well is 50 m.

(b) If we ignore the travel time for the sound, we can use the same equation to calculate the depth of the water well. We get x = (1/2)(-9.8)(3.2)^2 = 50.24 m. This is the same answer as before because the time it took for the sound to reach the person's ear is negligible compared to the time it took for the rock to hit the water. Therefore, the calculated depth of the water well is still 50 m.