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If in a projection motion, the maximum h...

If in a projection motion, the maximum height is equal to a horizontal range, find the angles of projection `[tan^(-1) (4) = 76^(@)]`

Text Solution

Verified by Experts

`H_(max) = R`
`(u^(2) sin^(2) theta)/(2 g) = (u^(2))/(g)(2 sin theta cos theta) rArr tan theta = 4`
`theta = tan^(-1)(4) = 76^(@)`
Another angle will be `90^(@) - tan^(-1)(4) = 14^(@)`.
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Knowledge Check

  • During a projectile motion if the maximum height equal the horizontal range, then the angle of projection with the horizontal is :

    A
    `tan^(-1)(1)`
    B
    `tan^(-1)(2)`
    C
    `tan^(-1)(3)`
    D
    `tan^(-1)(4)`

  • A projectile is projected from ground such that the maximum height attained by, it is equal to half the horizontal range.The angle of projection with horizontal would be

    A
    `tan^(-1) (2)`
    B
    `tan^(-1) (3)`
    C
    `tan^(-1) (4)`
    D
    `tan^(-1) sqrt(2)`

  • The horizontal range is four times the maximum height attained by a projectile. The angle of projection is

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

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