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A particle is projected from horizontal ...

A particle is projected from horizontal ground with speed `5ms^(-1)` at `53^(@)` with horizontal. Find time after which velocity of particle will be `45^(@)` with horizontal:

A

`(1)/(10)` sec

B

`(3)/(10)` sec

C

`(5)/(10)` sec

D

`(7)/(10)` sec

Text Solution

Verified by Experts

The correct Answer is:
a,d

`5cos 53^(@)=v cos 45^(@)`
`v sin 45^(@)=5sin 53^(@)-gt`
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