AP Physics Homework, help!?

Question

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 1.80 s after the stone is dropped. How deep is the hole?

*A bunch of people gave me wrong answers to this and as a result I only have 1 submission left* Please make sure the answer is correct, it is a two step thing I Believe.

Answer 1

http://answers.yahoo.com/question/index;_ylt=AshXzIJ3gErzDArRsvPdPwHty6IX?qid=20070820035431AAtHBMy&show=7#profile-info-LYwzvBaraa

Answer 2

The total time represents the time to fall plus the time for the sound to travel back up: T1 + T2.
Let a = acceleration due to gravity
s = speed of sound
let d be the unknown distance

we know d = .5*at^2 , so T1 = (2d/a)^.5

T2 is simply d/s

thus 1.80 sec = (2d/a)^.5 + d/s
a and s are known values so it's a simple matter to solve for d. I didn't solve it because you gave values in metic. Being from the old school I am accustomed to expressing a as 32ft/sec^2. Since I wasn't sure how close your answer has to be I leave it to you to plug in the values for a and s.
P.S.: If the form of the last equation gives you a problem then simply substitute K^2 for d, solve it as a binomial problem for K, and then take the sq root of K as the answer as d.

Answer 3

Draw a picture first. This should form a triangle from the plane straight down to the ground, up the hill, and then back horizontally to the plane. You already know one of the legs and from trig you can find out one of the angles. From that information you find out that horizontal part of the traingle (which is the distance the pilot will fly before he hits the ground. Once you find this distance, use the formula d=v*t with d=distance, v=velocity, t=time to solve for how long the pilot has until hitting the ground.

Answer 4

I answered one like this earlier....see resolved questions...it was for 3 seconds.

total time of falling rock and sound wave returning to your ear is 1.8 sec.
t = time for rock to hit....let x = time for sound to return
then t + x = 1.8 seconds
so x = 1.8 - t
h = 0.5gt^2 for the rock, and rate*time = dist sound must travel..........they must be equal.
thus 0.5gt^2 = 343*x, or 0.5gt^2 = 343(1.8-t)
simplify and solve the quadratic..........
use g = 9.8
4.9t^2 = 617.4-343t
4.9t^2+343t - 617.4 = 0
using TI graphic, I get t = 1.756 sec...............verify
h = 0.5 * 9.8 * ( 1.756)^2 = 15.109m
sound distance = 343 ( 1.8-1.756) = 343*0.044 = 15.092m [ if more decimal places are used, it will be closer...........both round to 15.1 m .....e.g....if t = 1.7559519, then both ans give 15.10849 m...........]

Answer 5

Let t1 = time falling
Let t2 = time for sound
Let s = depth of well
t1 + t2 = 1.80 sec.
s = 1/2 a * t1 ^2
s = 4.9 * t1^2
t1 = PosSquareRoot(s/4.9))
t2 = s /343
1.80 = PosSquareRoot(s/4.9)) + s/343
1.80 = PosSquareRoot(s) / 2.214 + s/343
Solve for s

Answer 6

velocity = distance / time

therefore 343 = distance / 1.8sec

transpose formula,

velocity x time = distance

343 x 1.8 = your answer