Starting from rest, your friend dives from a high cliff into a deep lake below,

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.

Explanation / Answer

(fo + ?f)/fo = c/(c - Vs) where c is the velocity of sound in the intervening medium. Substituting the given values fo = 919 Hz, ?f = 55.9 Hz and c = 344 m/s into this equation gives (919 + 55.9)/919 = 344/(344 - Vs) which may be rearranged to give the instantaneous velocity Vs = 344*55.9/(919 + 55.9) = 19.72 m/s Since the source was initially at rest, and is falling freely with a gravitational acceleration of 9.80 m/s², the time t required to reach this velocity is given by Vs = g.t that is, in a time t = Vs/g = 2.01 s.

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