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A particle is projected form a horizonta...

A particle is projected form a horizontal plane `(x-z` plane) such that its velocity vector at time `t` is gives by `vecV = ahati +(b - ct)hatj`. Its range on the horizontal plane is given by

A

`(ba)/c`

B

`(2ba)/c`

C

`(3ba)/c`

D

None

Text Solution

Verified by Experts

The correct Answer is:
B
`vecV = ahati + (b-"ct")hatj = 4xhati + (U_(y)-"gt")hatj`
`R = (2u_(x) U_(y))/g = (2ab)/c`
`v^(2) = 2g sin theta xx l`
`v^(2)/n^(2) = 2(g sin theta - mu g cos theta)l`
`sin theta(1-1/(1/n^(2))) = mu cos theta`
`mu = tan theta (1- 1/n^(2))`
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Knowledge Check

  • The velocity at the maximum height of a projectile is half of its velocity of projection u . Its range on the horizontal plane is

    A
    `(3 u^(2))/(g)`
    B
    `(sqrt(3) u^(2))/(2 g)`
    C
    `(3u^(2))/(g^(2))`
    D
    `(u^(2))/(3g)`

  • A particle is projected from the horizontal x-z plane, in vertical x-y plane where x-axis is horizontal and positive y-axis vertically upwards. The graph of y coordinate of the particle v/s time is as shown . The range of the particle is sqrt3 . Then the speed of the projected particle is:

    A
    `sqrt3(m)/(s)`
    B
    `sqrt((403)/(4))(m)/(s)`
    C
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  • A particle is projected from a horizontal plane with a velocity of 8sqrt2 ms ^(-1) ms at an angle theta . At highest point its velocity is found to be 8 ms^(-1) Its range will be (g=10 ms^(-2))

    A
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    B
    4.6 m
    C
    6.4 m
    D
    12.8 m

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