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

Answer

D

Explanation

`u cos 60^(@) =5, V_(y) =u sin 60^(@) -10t`
`V^(2) =(u sin 60^(@)-10t)^(2)+(u cos 60^(@))`
`(u^(2))/4=(u(sqrt(3))/2-10 t)^(2)+(u^(2))/4`
`rArr 10 t=(10sqrt(3))/2rArr t=(sqrt(3))/2`