rest and motion ; Kinematics?

Question

An NCC parade is going at a uniform speed of 6km/hr through a place under a berry tree on which a bird is sitting at a height of 12.1 m. At a particular instant a bird drops berry. Which cadet will receive the berry on his uniform?

Answer 1

haha! A berry, huh?
OK, the "dropping" will spend a certain amount of time in the air.
Let's say that the height is 12.1 meters ABOVE the spot on the uniform where the "berry" will splat.
The time the "berry" spends falling is found using s = 1/2at^2
or t^2 = 12.1/(1/2*9.8)
or t = sqrt(12.1*2/9.8) or 1.57 seconds.

So, how far has the parade traveled in 1.57 seconds?
6 km/hr = 6 kh/hr*1000 m/km*1hr/3600 seconds = 6000/3600 m/s = 1.666 m/s
or 1.666*1.57 meters total in the time it takes the "berry" to drop.
This is 2.6 meters.

The cadet who is 2.6 meters behind the bird when it "drops" the "berry" is the one who will be hit with it.