a stone is dropped from the top of building 200 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 50 metre per second. find where and when the stone will meet
Answers
Explanation:
Given: The height of the tower is 100 m and the initial velocity of the second stone is
25
m/s
Case-1: Stone is dropped from the tower.
The second equation of motion is given as,
s=ut+1/2gt²
where, the distance covered by stone is
s
,the initial velocity is
u
, the acceleration due to gravity is
g
, and the time taken is
t
when two stones meet.
We know that the value of acceleration due to gravity is
10m/s²
By substituting the given values in the above equation, we get,
s=ut+1/2gt²
s=0×t+1/2×10×t²
s=5t²
Case-2: Stone is thrown in upward direction.
The second equation of motion is given as,
s’=ut+1/2gt²
By substituting the given values in the above equation, we get,
s’=ut+1/2gt’
s’=50×t−1/2×10×t²
s’=25t−5t²
The combined displacement of both the stones at the meeting point is given as,
s+s’=200
By substituting the values of
s
and
s’
in the above equation, we get,
5t²+25t−5t²=100t=4s
The covered distance by falling stone is given as,
s=ut+1/2gt²
s=0+1/2×10×(4)²
s=80m
Thus, the stones will meet after 4s
at a height
80
m
from the top.