Free Fall(gravity) with speed of sound in air

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SUMMARY

The discussion revolves around calculating the time it takes for a pebble dropped into a well to hit the water, the time for the sound of the splash to reach the observer, and the depth of the well. The total time for both events is 3.0 seconds, with the speed of sound in air being 340 m/s. The pebble accelerates under gravity at -9.8 m/s², and the average speed during its fall is approximately 15 m/s. The calculations reveal that the majority of the 3 seconds is spent on the pebble's descent.

PREREQUISITES
  • Understanding of kinematic equations for uniformly accelerated motion
  • Knowledge of the speed of sound in air (340 m/s)
  • Basic concepts of gravity (g = -9.8 m/s²)
  • Ability to solve quadratic equations
NEXT STEPS
  • Calculate the time taken for the pebble to hit the water using kinematic equations
  • Determine the depth of the well using the derived time and average speed
  • Explore the effects of air resistance on falling objects
  • Learn about sound propagation in different mediums
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of free fall and sound propagation in air.

Patrick817
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A pebble is dropped into a deep well, and 3.0s later the sound of a splash is heard as the pebble reaches the bottom of the well. The speed of sound in air is 340m/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?

t=?
Initial Velocity=0
g=-9.8

I've used all of the formulas I have and still haven't been able to figure out how to get this. Someone please help. (At least with A)

Formulas Given: d=.5(final velocity + initial velocity)t
d= initialvelocity(t) + .5at^2
final velocity^2= initialvelocity^2+ 2ad
 
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Patrick817 said:
A pebble is dropped into a deep well, and 3.0s later the sound of a splash is heard as the pebble reaches the bottom of the well. The speed of sound in air is 340m/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?

t=?
Initial Velocity=0
g=-9.8

I've used all of the formulas I have and still haven't been able to figure out how to get this. Someone please help. (At least with A)

Formulas Given: d=.5(final velocity + initial velocity)t
d= initialvelocity(t) + .5at^2
final velocity^2= initialvelocity^2+ 2ad

The event can be divided into two parts:
First part, the stone accelerates to the bottom of the well where it makes a noise.
Second part, the sound travels back up the well at constant speed to the observer.
The total time is 3 seconds.

A stone that falls for 3 seconds reaches almost 30m/s, and so averages about 15m/s.
So the average speed for the first part is ~15m/s
The average speed for the second part is 340 m/s

So when you work it out, most of the 3 seconds will be the stone falling, and a little bit at the end will be the sound coming up.

Hope that helps.
 

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