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The horizontal range of a projectile is ...

The horizontal range of a projectile is times of its maximum height. Its angle of projection will be :-

A

`45^(@)`

B

`60^(@)`

C

`90^(@)`

D

`30^(@)`

Text Solution

Verified by Experts

The correct Answer is:
D
`R=(u^(2) sin 2 theta ) /(g) and H=(u^(2) sin ^(2) theta )/( 2g) `
but ,` R= 4 sqrt(3 ) H `
`(u^(2) sin 2 theta ) /(g) = 4 sqrt(3) ((u^(2) sin ^(2) theta )/(2g))`
` sin 2 theta = 4 sqrt(3)(sin^(2)theta )/(2)`
`2 sin theta cos theta =(4sqrt(3) sin ^(2)theta )/(2)`
` (1)/(sqrt(3))-(sin theta )/(cos theta ) = tan theta `
or `theta = 30^(@)`
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Knowledge Check

  • The horizontal range of a projectile is 4 sqrt(3) times its maximum height. Its angle of projection will be

    A
    `45^(@)`
    B
    `60^(@)`
    C
    `90^(@)`
    D
    `30^(@)`

  • The horizontal range of a projectile is 4sqrt(3) times its maximum height. The angle of projection is

    A
    `30^(@)`
    B
    `45^(@)`
    C
    `60^(@)`
    D
    None

  • The maximum horizontal range of a projectile is 400 m . The maximum value of height attained by it will be

    A
    `100 m`
    B
    `200 m`
    C
    `400 m`
    D
    `800 m`

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