Physics question about waves and freefall?

Question

A stone is dropped into a deep canyon and is heard to strike the bottom 10.2 seconds after release. The speed of sound waves in air at the canyon is 334 m/s. How deep is the canyon?

Method and answer would be much appreciated!

Answer 1

??

Answer 2

Okay, so if you heard the stone drop after 10.2 seconds, the actual drop time is less than that. Let's call the depth or height of the canyon H.

we'll use the vertical distance formula
height = (1/2)gt^2
H = (1/2) * 9.8 * t^2, where t is the amount of time it takes to fall.
you can solve for t and get:
t = sqrt(H/4.9)

The sound waves, when coming back up to you, must travel the height of the canyon, H. That would require an additional time of H/334 s.

So your total time, 10.2 = sqrt(H/4.9) + H/334
now, you must solve for H. You'll end up with a kinda nasty quadratic with a little bit of algebra. Just plug that into a solver and you will solve for your H, the depth of the canyon.

Answer 3

The distnace teh stone falls is:

y = -1/2g t where t = time for stone to fall

The sound travels the same distance according to:

y = vs ts and ts = time for sound to come back up and vs = speed of sound.

Now t + ts = T =10.2 seconds so

y = vs*(T-t)

Set the two equations for y equal to each other - it is the same distance

vs*(T-t) = 1/2 g t^2 Solve for t

0 = 1/2 g t^2 + vs*t -vsT

Use quadratic formula:

t = -vs/2 +/- 1/2*sqrt(vs^2 +2gvs)

The using only the positive value for t, find y by

y = vs*(T-t)

I'll leave teh number crunching for you