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.

Options

A. `(1)/(2)sec`
B. `1sec`
C. `sqrt(3//2)sec`
D. `sqrt(3//2)sec`

Answer

D

Explanation

`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)`.