physics
posted by shaknocka on .
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?

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.