In the time taken by the projectile to reach from `A` to `B` is `t`. Then the distance `AB` is equal to.
.

Options

A. `(ut)/(sqrt3)`
B. `(sqrt3ut)/(2)`
C. `(sqrt3ut)`
D. `2ut`

Explanation

`(V_(o))_(x)=ucos30^(@),a_(x)=-gsin30^(@)`
`AB=(ucos30^(@))t-(1)/(2)gsin30.t^(2)`
`AB=(ucos30^(@)).t-(1)/(2)gsin30^(@).t^(2)`
`AB=(sqrt3ut)/(2)-(g t^(2))/(4)`
but `t=(2usin(60^(@)-30^(@)))/(gcos30^(@))=(2u)/(sqrt3g)`
`:.AB=(sqrt3ut)/(2) -(g t)(t)/(4)`
`=(sqrt3ut)/(2)-((2u)/(sqrt3))(t)/(4)`
`=(ut)/(2)(sqrt3-(1)/(sqrt3))=(ut)/(sqrt3)`
Altrnate method : horizontal distance in time `t=AC=ucos60^(@).t`
`AC=(ut)/(2)`
and `AB=(AB)/(cos30^(@))=(ut//2)/(sqrt3//2)=(ut)/(sqrt3)`