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Two particles are projected simultaneously from two points `O` and `O'` such that `10 m` is the horizontal and `5 m` is the vertical distance between them as shown in the figure. They are projected at the same inclination `60^@` to the horizontal with the same velocit `10 ms^(-1)`. The time after which their separation becomes minimum is

A

`2.5s`

B

`1s`

C

`5s`

D

`10s`

Text Solution

Verified by Experts

The correct Answer is:
B
Let us find the relative velocity of `O`. w.r.t `O'`.
`vecvO//(O')=vecvO-vecvO'=(10cos60^@hati+10sin60^@hatj)`
`-[10cos60^@(-hati)+10sin60^@hatj]=10hati`
This is along the horizontal direction.

If we assume `O'` to be at rest, then `O` will move along `OA` w.r.t `O'` and the minimum separation will be when particle `O` is at `A`.
For this the relative displacement travelled `=OA=10 m`
Time taken `=10/(V_(O)//O')=10/10=1s`
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