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Two bodies are thrown simultaneously fro...

Two bodies are thrown simultaneously from a tower with same initial velocity `v_0`: one vertically upwards, the other vertically downwards. The distance between the two bodies after time t is

A

`2V_0t+1/2 "gt"^2`

B

`2v_0t`

C

`v_0t+1/2 "gt"^2`

D

`v_0t`

Text Solution

Verified by Experts

The correct Answer is:
B
For vertically upward motion, `h_1 = v_0t -1/2 " gt"^2` and for vertically down ward motion, `h_2 = v_0t-1/2 "gt"^2`
`:. " Total distance covered in " t sec h= h_1+ h_2 = 2v_0t`
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