By Team Aakash Byju's | 31st January 2023
2v t+ gt
Two bodies are thrown simultaneously from a tower with the same initial velocity v0 one vertically upwards, the other vertically downwards. The distance between the two bodies after time t is.
-1
o
2
1
2
2v t
o
v t+ gt
o
1
2
v t
o
Detailed Explanation
o
Given that one body is thrown vertically upwards & another body is thrown vertically downwards from a tower of height say h with the same initial velocity v .
Let s be the distance of the body from the tower which is thrown upward and s be the distance of the body from the tower which is thrown downward.
1
2
v
v
o
o
2
1
g
Thus distance between two bodies is:
=> s =s +s
1
2
From the second equation of motion, we can write
v t- gt
o
1
2
v t+ gt
o
1
2
s =
1
s =
2
For vertically upward motion
2
2
For vertically downward motion
Total distance covered in t sec
=> s =s +s
1
2
= 2v t
o
. .
.