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Patrick817 Feb15-12 06:22 PM

Free Fall(gravity) with speed of sound in air
 
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 havent 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

PeterO Feb15-12 06:59 PM

Re: Free Fall(gravity) with speed of sound in air
 
Quote:

Quote by Patrick817 (Post 3766408)
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 havent 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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