Posted by **jean** on Thursday, October 23, 2008 at 1:20pm.

A spelunker drops a stone from rest into a hole. The speed of sound is 343m/s in air, and the sound of the stone striking the bottom is heard 1.46s after the stone is dropped. How deep is the hole? What is it in meters

- physics -
**bobpursley**, Thursday, October 23, 2008 at 1:24pm
time=time down + time up

time=sqrt(2h/g)+ h/vsound

solve for time Use the quadratic equation.

- physics -
**drwls**, Thursday, October 23, 2008 at 1:28pm
Solve the equation

t1 + t2 =

(stone's time required to fall) + (time to hear the splash) = 1.46 s

(1/2) g t1^2 = H

t1 = sqrt (2H/g)

a H = t2 (a is the sound speed, 343 m/s)

t2 = H/a

You will have to set it up as a quadratic equation in H, the depth of the well.

- physics -
**Anonymous**, Tuesday, December 4, 2012 at 10:28pm
10.24

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