(2-1)   ①  Using Newton’s equation of motion, s = ut + ½at2, to describe the downward motion:    

u is zero, so s = ½at2. Now s is the distance to the ground, which is just 1.2m, and hence the time taken for the shot to land is:

                                                          t = √(2s/a)  =  √(2 x 1.2 / 9.8)  =  0.50s

The velocity in the forward direction is 10 ms-1, so we can simply calculate the throw distance as the distance travelled in this direction during this 0.50 s.

Hence, s = ut  =  10 x 0.50  = 5.0 m

②  From Nefertiti’s point of view, the situation is no different from when she was on the ground, so she expects that her throw will still reach the same distance, 5.0 m.

③  The overall velocity of the shot is the plane’s velocity plus the additional velocity added by Nefertiti:  so it is 150 + 10  =  160 ms-1.  The time taken to fall to the floor (of the cabin, this time) is still 0.5s. Therefore the distance travelled by the shot towards Cormorant is:

ut  =  160 x 0.5  =  80m

④  The plane’s velocity is just 150 ms-1, so during the 0.5s during which the shot is moving, it will have moved 150 x 0.5 = 75m

⑤  The distance travelled by the shot, relative to that travelled by the cabin floor, is therefore 80 – 75 = 5.0m.   In other words, the different frame of reference makes no difference to the outcome – Nefertiti’s throw measured within the plane is still predicted to be the same.