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

. .

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