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For a projectile 'R' is range and 'H' is...

For a projectile 'R' is range and 'H' is maximum height
`{:("List-I", "List-II"),((A)R=H, (e) "Angle of projection" tan^(-1)(1)),((b)R=2H, (f)"Angle of projection" tan^(-1)(4)),((c)R=3H,(g)"Angle of projection" tan^(-1)(2)),((d)3=H,(h)"Angle of projection" tan^(-1)(4//3)):}`

A

a-g, b-h, c-e, d-f

B

a-h, b-g, c-e, d-f

C

a-f, b-g, c-h,d-e

D

a-e, b-g, c-f, d-h

Text Solution

Verified by Experts

The correct Answer is:
C
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The horizontal range of a projectile is 2sqrt(3) time its maximum height. Find the angle of projection .

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Knowledge Check

  • tan h^(-1)((1)/(2)) =

    A
    `(1)/(2) log_(e)3`
    B
    `(1)/(2) log_(e)2`
    C
    `log_(e)3`
    D
    `log_(e)5`

  • tan h^(-1) ((1)/(3))

    A
    `(1)/(2) log_(e) (2)`
    B
    `2 log_(e) (2)`
    C
    `-(1)/(2) log_(e) (2)`
    D
    `-2 log_(e) (2)`

  • tan h^(-1) ((1)/(2)) =

    A
    `(1)/(2) log_(e) 3`
    B
    `(1)/(2) log_(e) 2`
    C
    `log_(e) 3`
    D
    `log_(e) 5`

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