In the time taken by the projectile to reach from `A` to `B` is `t`. Then the distance `AB` is equal to. .A. `(ut)/(sqrt3)`B. `(sqrt(3) ut)/2`C. `sqrt3 ut`D. `2ut`

In the time taken by the projectile to reach from `A` to `B` is `t`. T

1.	In the time taken by the projectile to reach from `A` to `B` is `t`. Then the distance `AB` is equal to. .A. `(ut)/(sqrt3)`B. `(sqrt(3) ut)/2`C. `sqrt3 ut`D. `2ut`
Answer» Correct Answer - a Horizontal component of velocity, `u_(H)=ucos 60^(@)=u/2` `AC=u_(H)xxt=(ut)/2` And `AB=AC sec 30^(@)` n `=((ut)/2)(2/(sqrt(3)))=(ut)/(sqrt(3))`

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In the time taken by the projectile to reach from `A` to `B` is `t`. Then the distance `AB` is equal to. .A. `(ut)/(sqrt3)`B. `(sqrt(3) ut)/2`C. `sqrt3 ut`D. `2ut`

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