Question

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?

Answer 1

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:
$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ \Rightarrow 12.1=0+\frac{1}{2}\times 9.8t^2 \\ \Rightarrow t^2=\frac{12.1}{4.9}=2.46 \\ \Rightarrow t=1.57\mathrm{s} \end{gathered}$$

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.