Question

Two bodies are thrown simultaneously from a tower with same initial velocity : 0 v one vertically upwards, the other vertically downwards. The distance between the two bodies after time t is (a) 2 0 2 1 2v t  g t (b) v t 2 0 (c) 2 0 2 1 v t  g t (d) v t

Answer 1

Answer:

For vertically upward motion,

h

1

=v

0

t−

2

1

gt

2

and for vertically downward motion,

h

0

=v

0

t+

2

1

gt

2

∴ Total distance covered in t sec

h=h

1

+h

2

=2v

0

t

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