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A particle is projected with speed 10 m/...

A particle is projected with speed `10 m//s` at angle `60^(@)` with the horizontal. Then the time after which its speed becomes half of initial.

A

`(1)/(2)sec`

B

`1sec`

C

`sqrt(3//2)sec`

D

`sqrt(3//2)sec`

Text Solution

Verified by Experts

The correct Answer is:
D
`u cos60^(@)=5,V_(y)=usin 60^(@)-10t`
`V_(2)=(u sin 60^(@)-10t)_(2)+(u cos 60^(@))`
`(u^(2))/(4)=(u(sqrt(3))/(2)-10t)^(2)+(u^(2))/(4)`
`rArr 10t=(10 sqrt(3))/(2) rArr t=(sqrt(3))/(2)`.
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