Pebble being dropped into a well

  • Thread starter professordad
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In summary, the pebble takes 3.0 seconds to hit the water and the sound takes 3.0 seconds to reach the observer. The well is 4.9 meters deep.
  • #1
professordad
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Homework Statement



A pebble is dropped into a deep well, and 3.0 seconds later the sound of a splash is heard as the pebble reaches the bottom of the well. The speed of sound in air is about 340 m/s.
(a) How long does it take for the pebble to hit the water?
(b) How long does it take for the sound to reach the observer?
(c) What is the depth of the well?

From College Physics by Serway and Faughn

Homework Equations



[itex]v = v_0 + at[/itex]

[itex]x = v_0t + \frac{1}{2}at^2[/itex]

[itex]v^2 = v_0^2 + 2ax[/itex]

The Attempt at a Solution



(a) So in this problem we know the acceleration (which is the gravitational constant) and we also know [itex]v_0[/itex] which is 0 because it starts out as 0 m/s. But in all three equations, there are 4 variables, and knowing 2 variables is not enough.

Then I tried plugging in and obtaining a system. So I got

[itex]-19.6x = v^2[/itex]
[itex]v = -9.8t[/itex]

But this doesn't help much, and if you add a 3rd equation its going to be the same thing (the three equations are related). So I don't know how to proceed :(

Also for (b) and (c) I think you need (a) to solve them.

BTW I'm out of school, so this problem doesn't count for a grade or anything :)
 
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  • #2
Forget the distance for the moment and try to find the time. You need the speed of sound for this.
 
  • #3
professordad said:

Homework Statement



A pebble is dropped into a deep well, and 3.0 seconds later the sound of a splash is heard as the pebble reaches the bottom of the well. The speed of sound in air is about 340 m/s.
(a) How long does it take for the pebble to hit the water?
(b) How long does it take for the sound to reach the observer?
(c) What is the depth of the well?

From College Physics by Serway and Faughn

Homework Equations



[itex]v = v_0 + at[/itex]

[itex]x = v_0t + \frac{1}{2}at^2[/itex]

[itex]v^2 = v_0^2 + 2ax[/itex]

The Attempt at a Solution



(a) So in this problem we know the acceleration (which is the gravitational constant) and we also know [itex]v_0[/itex] which is 0 because it starts out as 0 m/s. But in all three equations, there are 4 variables, and knowing 2 variables is not enough.

Then I tried plugging in and obtaining a system. So I got

[itex]-19.6x = v^2[/itex]
[itex]v = -9.8t[/itex]

But this doesn't help much, and if you add a 3rd equation its going to be the same thing (the three equations are related). So I don't know how to proceed :(

Also for (b) and (c) I think you need (a) to solve them.

BTW I'm out of school, so this problem doesn't count for a grade or anything :)
How have you defined your variables?

I don't see any of your solution which includes the speed of sound.

Don't assume that you can get part (a) without also working on (b) and/or (c) .
 
  • #4
Taking the depth to be d, the time it takes for the rock to hit the water is [itex]t_1= \sqrt{d/4.9}[/itex] and the time for the sound to come back is [itex]t_2= d/340[/itex]. So the total time is given by [itex]t_1+ t_2= \sqrt{d/4.9}+ d/340= 3[/itex].
 
  • #5
@ SammyS: Oops yeah, I did forget to define variables. In this case [itex]v_0[/itex] is the starting velocity, [itex]t[/itex] is the time which it takes for the pebble to hit the water, and [itex]x[/itex] is the displacement. (in this case it's also the distance because the pebble doesn't go back) And [itex]v[/itex] is the velocity when the pebble hits the water, though I guess that wasn't needed.

Thanks to everyone for the help! :D
 

1. How deep is the well that the pebble is being dropped into?

The depth of the well can vary depending on the location and type of well. It is important to know the depth in order to calculate the time it takes for the pebble to hit the bottom and the force of impact.

2. What is the weight of the pebble being dropped?

The weight of the pebble will also affect the speed and force of impact when dropped into the well. This can be measured using a scale or estimated based on the size and type of pebble.

3. How does the speed of the pebble change as it falls into the well?

The speed of the pebble will increase as it falls due to the force of gravity. However, it will eventually reach a maximum speed, known as terminal velocity, when the force of air resistance equals the force of gravity.

4. What factors can affect the trajectory of the pebble as it falls into the well?

The trajectory of the pebble can be affected by factors such as wind, air resistance, and any obstacles in the path of the pebble. These factors can cause the pebble to deviate from a straight drop and potentially change the force of impact.

5. Can the depth of the well affect the sound of the pebble hitting the bottom?

Yes, the depth of the well can affect the sound of the pebble hitting the bottom. As the pebble falls, the sound waves created by the impact will travel back up the well. The deeper the well, the longer it will take for the sound waves to reach the top, resulting in a longer echo or reverberation.

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