An NCC parade is travelling at a uniform speed of 6 km/h through a...

Speed of the cadet, $v = 6 \text{ km/h} = 6 \times \dfrac{5}{18} \text{ m/s} = \dfrac{5}{3} \text{ m/s}$

Height of the bird, $h = 12.1 \text{ m}$

The initial velocity of the dropped berry is $u = 0$.

Using the equation of kinematics for the vertical motion of the berry:

$h = ut + \dfrac{1}{2}gt^2$

$12.1 = 0 + \dfrac{1}{2}(9.8)t^2$

$12.1 = 4.9t^2$

$t^2 = \dfrac{12.1}{4.9} = \dfrac{121}{49}$

$t = \dfrac{11}{7} \text{ s}$

The distance of the cadet from the tree at the instant the berry was dropped is equal to the distance travelled by the cadet during the time $t$.

$d = v \times t = \dfrac{5}{3} \times \dfrac{11}{7} = \dfrac{55}{21} \text{ m}$

$d \approx 2.62 \text{ m}$