Question

Starting from rest, your friend dives from a high cliff into a deep lake below, yelling in excitement at the thrill of free-fall on her way down. You watch her as you stand on the lake shore immediately below, and at a certain instant your keen hearing recognizes that the usual frequency of her yell, which is 903 Hz, is shifted by 53.1 Hz.
(A) How long has your friend been in the air when she emits the yell whose frequency shift you hear? Take 342 m/s for the speed of sound in air and 9.80 m/s^2 for the acceleration due to gravity.

Answer 1

Answer:

Time spent = 1.88 sec.

Explanation:

This is a case of doppler shift of a moving source and a stationary receiver.

f' = cf/(c - u)

Where f' is the apparent frequency = 903 + 51.3 = 954.3 Hz

c = speed of sound = 342 m/s

f = real frequency = 903 Hz

u = speed of sound source = ?

Solving for u we have

954.3 = (342 x 903)/(342 - u)

954.3 = 308826/(342 - u)

326370.6 - 954.3u = 308826

-954.3u = 308826 - 326370.6

-954.3u = -17544.6

u = 18.38 m/s

Using u = u° + gt

Where u = velocity of your friend at that instance = 18.38 m/s

u° = your friend's initial velocity = 0 (your friend started falling from rest)

g = acceleration due to gravity 9.8 m/s^2

t = time spent in the air by your friend.

18.38 = 0 + 9.8t

t = 18.38/9.8 = 1.88 sec.