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Time taken by the particle to reach from...

Time taken by the particle to reach from `A` to `B` is `t`. Then the distance `AB` is equal to

A

`(ut)/(sqrt(3))`

B

`(sqrt(3)ut)/2`

C

`sqrt(3)ut`

D

`2 ut`

Text Solution

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The correct Answer is:
A
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Knowledge Check

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

    A
    `(ut)/(sqrt(3))`
    B
    `(sqrt(3) ut)/(2)`
    C
    `sqrt(3) ut`
    D
    `2 ut`

  • If time taken by the projectile to reach Q is T , than PQ =

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    `Tv sin theta`
    B
    `Tv cos theta`
    C
    `Tv sec theta`
    D
    `Tv tan theta`

  • A particle is dropped from point A at a certain height from ground. It falls freely and passes through thre points B, C and D with BC=CD . The time taken by the particle to move from B to CD is 2s and from C to D is 1s . The time taken to move from A to B is s

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    C
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    D
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