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/s2
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.
