Posted by **shaknocka** on Monday, February 7, 2011 at 10:51am.

A spelunker (cave explorer) drops a stone from rest into a hole. The speed of sound is 343 m/s in air, and the sound of the stone striking the bottom is heard 3.24 s after the stone is dropped. How deep is the hole?

- physics -
**Henry**, Tuesday, February 8, 2011 at 9:32pm
t1 + t2 = 3.24s.

t2 = 3.24 - t1.

t1 = Flight time of stone.

t2 = Travel time of the sound.

d1 = 0.5gt^2 = distance the stone traveled.

d2 = Vt = Distance the sound traveled.

d1 = d2,

0.5*9.8t^2 = 343m/s * t2,

Substitute 3.24 - t1 for t2:

4.9t1^2 = 343 * (3.24 - t1),

4.9t1^2 = 1111.3 - 343t1,

4.91t1^2 + 343t1 - 1111.3 = 0,

Use Quadratic Formula to find t and get:

t1 = 3.10244, and -73.10.

Use the positive value of t.

t1 = 3.10244s = time in flight for stone.

d1 = 0.5 * 9.8 * (3.10244)^2,

d1 = 47.2m = Depth of hole.

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