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. Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

Solution:

Let t be the true time of fall 
Initial velcoity=u=0m/s
a=9.8 m/s2
Distance=S=12.1 m
S= ut+1/2 at²
12.1=o+1/2 x 9.8x t²
t²=12.1/4.9=2.46 sec
t=1.57 sec
Cadet velocity=6km/hr=6x5/18=1.66 m/s

Distance = Speed x time
=1.57 x 1.66 =2.6m
The cadet , at 2.6m away from the tree will receive the berry on his uniform.

Answer 2

Answer:2.62m

Explanation:speed of the person is 6km/h

s=1/2gt*t

12.1m=1/2*9.8m/s*t*t

by caluclating this we will get=

t*t=121/49

so,t=11/7sec

convert 6 km/h into m/s

6*5/18=5/3m/s

so,distance=speed*time

d=5/3*11/7

d=2.6190.....

approximately=2.62