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Find the angle of projection of a porjec...

Find the angle of projection of a porjectile for which for horizontal range and maximum height are equal.

Text Solution

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Given, `R = H`
`:. (u^(2) sin 2 theta)/(g) = (u^(2) sin^(2) theta)/(2g)`
or `2 sin theta cos theta = (sin^(2) theta)/(2)`
or `(sin theta)/(cos theta) = 4` or `tan theta =4`
`:. Theta = tan^(-1)(4)`
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Knowledge Check

  • Find the angle of projection of a projectile for which the horizontal range and maximum height are equal.

    A
    `alpha = tan^-1(2)`
    B
    `alpha = tan^-1(4)`
    C
    `alpha = tan^-1(5)`
    D
    None of the above

  • The angle of projection at which the horizontal range and maximum height of projectile are equal is

    A
    `45^@`
    B
    `theta = tan^(-1) (0.25)`
    C
    `theta = tan^(-1) (4)`
    D
    `60^(@)`

  • If the angle of projection theta corresponds to horizontal range being equal to the maximum height then tantheta equals :

    A
    `1`
    B
    `1//sqrt(3)`
    C
    `sqrt(3)`
    D
    `4`

    • Similar Questions

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    The angle of rojection for a body to have same horizontal range and maximum height is …………………… .

    For angles of projection 15° and 45°, ratio of horizontal ranges will:

    A body is thrown at an angle alpha to the horizontal with initial velocity v_0 .Assuming air drag to be negligible , find (i)the time of motion, (ii)maximum height of ascent and the horizontal range, (iii)the angle of projection at which they are equal , (iv)the angle of projection at which the horizontal range is maximum and the angle at which the vertical range is maximum.

    For which angle of projection the horizontal range is 5 times the maximum height attrained ?

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