Question

1. A rock is thrown downward into a well that is 8.70 m deep.

If the splash is heard 1.10 seconds later, what was the initial speed of the rock? Take the speed of sound in the air to be 343 m/s.

2. A bird-watcher is hoping to add the white-throated sparrow to her "life list" of species. Person, only 1.00 m from the bird, hears the sound with an intensity of 2.68×10^?5 W/m² .

How far could she be from the bird and still be able to hear it? Assume no reflections or absorption of the sparrow's sound.

3. At Zion National Park a loud shout produces an echo 2.00 s later from a colorful sandstone cliff.

How far away is the cliff? Take the speed of sound in the air to be 343 m/s .

Answer 1

1) Let h = height = 8.70 m
S = speed of sound =343 m/s
T = Total time = 1.10 sec
t1 = the time to drop
t2 = the time for the sound to come back
t2 = h / S
T = t1 + t2
t1 = T - t2
t1 = T - (h / S)
t1 = 1.10 sec - (8.70m / 343 m/s)
t1 = 1.075 sec
Now use a free fall equation to find V0 as a function of t1
h = V0 * t1 + (1/2) g * t1^2
V0 * t1 = h - (1/2) g * t1^2
V0 = (h - (1/2) g * t1^2) / t1
V0 = (- 8.7 m - (1/2) (-9.81 m/s^2) * (1.075 sec)^2)/(1.075sec)
V0 = -2.82 m/s

2) The sound spreads out according to the inverse square law. This means the intensity is inversely proportional to distance squared:
I ? 1/d²
As an equation, this is
I = k/d²

when d=1.00m, I = 2.68×10?? W/m².

So k = Id² = 2.680×10?? x 1.00² = 2.68×10?? W
(In fact k is the power output of the bird.)
The limit of human hearing is usally taken as an intensity of 10?¹² W/m². The distance that gives this intensity is calculated as follows:
I = k/d²
10?¹² = 2.680×10?? / d²
d² = 2.680x10?
d = 5176.87m

d = 5180 m ( two significant figures)

3) 2d = vt

v = 343m/s , t=2.00 sec
d = vt/2 = 343*2.00/2 = 343 m