Ozone in troposphere is also know as Bad Ozone, Evil Ozone and Ground Level Ozone.
Answer:
A) 1568.60 Hz
B) 1437.15 Hz
Explanation:
This change is frequency happens due to doppler effect
The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source

where
C = the propagation speed of waves in the medium;
Vr= is the speed of the receiver relative to the medium,(added to C, if the receiver is moving towards the source, subtracted if the receiver is moving away from the source;
Vs= the speed of the source relative to the medium, added to C, if the source is moving away from the receiver, subtracted if the source is moving towards the receiver.
A) Here the Source is moving towards the receiver(C-Vs)
and the receiver is standing still (Vr=0) therefore the observed frequency should get higher

B)Here the Source is moving away the receiver(C+Vs)
and the receiver is still not moving (Vr=0) therefore the observed frequency should be lesser

You didn't attach the figure. Your text is incomplete. And you never got around to asking a question. Other than that, we're on it.
-- She went up for 0.4 sec and down for 0.4 sec.
-- The vertical distance traveled in gravity during ' t ' seconds is
D = (1/2) x (g) x (t)²
= (1/2) (9.8 m/s²) (0.4 sec)²
= (4.9 m/s²) x (0.16 s²)
= 0.784 meter ( B )
The impulse imparted to the shells equals the change in the momentum:
Fav*(Delta t)= Delta m*v.
The mass change is
Delta m= n*m= (89.9shells)*(88.7g)=7.97Kg
So the average force is
F=((v)*(Delta m))/t= ((929)*(7.97))/4.84=1529.78 N
Since the velocity of the shells is much greater than the velocity of the helicopter, there is no need to use relative velocity.