Please answer this.... No one can answewr it in our class>>> even our Laboratory and substitute teacher..?

Question

A rock is dropped down a dark well and a splash is heard 3.000 sec later. Taking into account the time required for the sound to travel up the well, calculate the distance to the water in the well. The speed of sound is 340 m/s... (the answer to this is 40. 65 but we can't find the solution.)

Answer 1

There are two times involved in the 3 seconds - the time the rock travels down to the water plus the time it takes for the sound to travel 2back up the well.

The distance traveled by a free falling object can be expressed as:

s = 1/2 g*t²

where

s = distance traveled by the object, g = 9.81 m/s² and t² is time squared

http://www.engineeringtoolbox.com/accela...

So the distance on the way down is (1/2)*9.81m/s²*t²

The distance on the way up is 340m/s *(3-t)   (if the rock is in the air for t seconds on the way down, the sound takes (3-t) seconds to travel up)

The distance down equals the distance up, so set them equal

(1/2)*9.81m/s²*t²= 340m/s *(3-t)

First, let's drop the units and know that the answer will be in seconds

(1/2)*9.81*t²= 340*(3-t)

4.905t²=1020-340t

4.905t²+340t-1020=0

Use the quadratic formula to solve (I just plugged in the coefficients into http://www.1728.com/quadratc.htm)

t=2.8803 or t=-72.197

Throw out the negative answer, we're not going to mess with negative time.

Now, what we solved for was the time on the way down.  Plug that into the first distance equation:

s=(1/2)*9.81m/s²*(2.8803)²=40.731

Hm.  

Plug in 3-t to the second distance equation:

340*(3-2.8803)=s=40.698

Well....close enough to call that a rounding error?

_/