Acar approaching a stationary observer emits 450. hz from its horn. if the observer detects a frequency pf 470. hz, how fast is the car moving? the speed of sound is 343 m/s.

The relationship between the frequency heard by the observer in motion and the original frequency of the sound is given by (Doppler effect)

where
f' is the frequency heard by the observer
v is the speed of the wave (the speed of sound)
is the speed of the source relative to the observer (= the speed of the car), and it is negative when the source is approaching the observer
f is the original frequency of the sound

By re-arranging the formula, we get

and by plugging the data of the problem into the equation, we find

so, the car is approaching the observer at 15.2 m/s.

The answer is as follows in the photo

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