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Equation of trajector of ground to groun...

Equation of trajector of ground to ground projectile is `y=2x-9x^(2)`. Then the angle of projection with horizontal and speed of projection is : `(g=10m//s^(2))`

A

`tan^(-1)(2),(5)/(2)m//s`

B

`tan^(-1)(3),(5)/(3)m//s`

C

`tan^(-1)(2),(2)/(3),(2)/(3)m//s`

D

`tan^(-1)(3),(2)/(3)m//s`

Text Solution

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The correct Answer is:
A
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Knowledge Check

  • The equation of trajectory of an oblique projectile y = sqrt(3) x - (g x^(2))/(2) . The angle of projection is

    A
    `30^(@)`
    B
    `45^(@)`
    C
    `60^(@)`
    D
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    `60^(@)`
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    D
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    A
    `sin^(-1)(4/5)`
    B
    `sin^(-1)(3/5)`
    C
    `sin^(-1)(4/3)`
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    In the following questions a statement of assertion (A) is followed by a statement of reason ( R). A : In the case of ground to ground projection of a projectile from ground the angle of projection with horizontal is theta =30^(@) . There is no point on its path such that instantaneous velocity is normal to the initial velocity . R : Maximum deviation of the projectile is 2theta = 60^(@) .

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