Posted by **Einstein** on Thursday, December 6, 2007 at 9:47pm.

Carol drops a stone into a mine shaft 12.6 m deep. How soon after she drops the stone does she hear it hit the bottom of the shaft?

- Physics -
**DrBob222**, Thursday, December 6, 2007 at 9:52pm
distance = 1/2 g*t^2

- Physics -
**Einstein**, Thursday, December 6, 2007 at 9:57pm
The mine shaft is 122.6 m deep.

How do I find time?

- Physics -
**~christina~**, Thursday, December 6, 2007 at 10:33pm
**The mine shaft is 122.6 m deep.
**

How do I find time?

you have the distance so..

12.6m or 122.6m (you changed it or you had a typo) but the set up is the same.

the equation Dr.Bob gave was correct but to elaborate on it:

Y(final height)= Y(initial height) + v(initial)t + 1/2 a(acceleration) t^2

is where he got the

distance= 1/2g*t^2

so since acceleration in y direction (or toward earth) is 9.81m/s^2 you use that as g or gravity in Dr.Bob's given equation or as "a" in my equation.

so now plugging in...(you only have to rearrange the equation to solve for t.)

sqrt[2(distance)/g] = t

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