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

#### Solution

Speed of the NCC cadets = 6 km/h = 1.66 m/s

Distance of the bird from the ground, *s* = 12.1 m

Initial velocity of the berry dropped by the bird, *u* = 0

Acceleration due to gravity, *a* = *g* = 9.8 m/s^{2}

Using the equation of motion, we can find the time taken *t* by the berry to reach the ground.

Thus, we have:

\[s = ut + \frac{1}{2}a t^2 \]

\[ \Rightarrow 12 . 1 = 0 + \frac{1}{2} \times 9 . 8 t^2 \]

\[ \Rightarrow t^2 = \frac{12 . 1}{4 . 9} = 2 . 46\]

\[ \Rightarrow t = 1 . 57 s\]

Distance moved by the cadets = *v* × *t* = 1.57 × 1.66 = 2.6 m

Therefore, the cadet who is 2.6 m away from tree will receive the berry on his uniform.