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A particle is projected up with a veloc...

A particle is projected up with a velocity of `v_(0)=10m//s` at an angle of `theta_(0)=60^(@)` with horizontal onto an inclined plane. The angle of inclination of the plane is `30^(@)`. The time of flight of the particle till it strikes the plane is (take `g=10ms^(-2)`)

A

`1s`

B

`1//2s`

C

`2/(sqrt(3))s`

D

`1/(2sqrt(3))s`

Text Solution

Verified by Experts

The correct Answer is:
c
`T=(2u_(y))/(a_(y))=(2v_(0)sin 30^(@))/(g cos 30^(@))=2/(sqrt3)s`
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Knowledge Check

  • A particle is projected with velocity of 10m/s at an angle of 15° with horizontal.The horizontal range will be (g = 10m/s^2)

    A
    10m
    B
    5m
    C
    2.5m
    D
    1m

  • a particle is projected down on inclined plane with a velocity of 21m/s at an angle of 60^@ with the horizontal its range on the inclined plane inclined at an angle of 30^@ with the horizontal is

    A
    21dm
    B
    2.1dm
    C
    30dm
    D
    6dm

  • a particle is projected down on inclined plane with a velocity of 21m/s at an angle of 60^@ with the horizontal its range on the inclined plane inclined at an angle of 30^@ with the horizontal is

    A
    21dm
    B
    2.1dm
    C
    30dm
    D
    6dm
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