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When the bomb is dropped, it will have an initial horizontal velocity which is equal to the speed of the aeroplane. So the bomb fall and travel forward too.

initial horizontal velocity,v_x = 600 km/h
v_x=600\ km/h = 600* \frac{5}{18}\ m/s=166.67\ m/s
initial vertical velocity,v_y = 0
height, h = 1960m
time taken to fall = t

h=v_yt+ \frac{1}{2}gt^2\\ \\1960=0+ \frac{1}{2}*9.8*t^2\\ \\1960=4.9t^2\\ \\t^2= \frac{1960}{4.9}=400\\ \\t= \sqrt{400}=20s

In 20s, the horizontal distance travelled is AB

AB=v_x*t=166.67*20=\boxed{3333.4\ m}
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