Afire engine is rapidly approaching the corner. explain what happens to the frequency of the sound wave as the fire engine draws closer. what do you hear?

My answer -

You will observe a higher apparent frequency because as the source moves closer relative to the observer the wavefront emitted will be closer to the wavefront previously emitted therefore shorter wavelength (in front of the source only) and since the speed of sound in air is assumed constant the frequency must increase.

P.S

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Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward