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, and at a certain instant your keen hearing recognizes that the usual frequency of her yell, which is 919 Hz, is shifted by 55.9 Hz. 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/s2 for the acceleration due to gravity.

Answer 1

Answer:

2 seconds

Explanation:

f = Frequency of yell = 919 Hz

$\Delta f$ = Shifted frequency = 55.9 Hz

v = Speed of sound in air = 342 m/s

$v_r$ = Velocity of friend

a = Acceleration due to gravity = 9.81 m/s²

From the Doppler shift formula we have

$\dfrac{f+\Delta f}{f}=\dfrac{v}{v-v_r}\\\Rightarrow v_r=v-\dfrac{vf}{f+\Delta f}\\\Rightarrow v_r=342-\dfrac{342\times 919}{919+55.9}\\\Rightarrow v_r=19.61\ m/s$

The velocity of the my friend is 19.61 m/s

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{19.61-0}{9.8}\\\Rightarrow t=2\ s$

The time my friend is in the air is 2 seconds