Home
Class 11
PHYSICS
The trajectory of a projectile in a vert...

The trajectory of a projectile in a vertical plane is `y=sqrt(3)x-2x^(2)`. `[g=10 m//s^(2)]`

Angle of projection `theta` is :-

A

`30^(@)`

B

`60^(@)`

C

`45^(@)`

D

`sqrt(3)` rad

Text Solution

Verified by Experts

The correct Answer is:
B
`y=sqrt(3)x-2x^(2)`
Trajectory equation is `y=x tan theta-(gx^(2))/(2u^(2) cos^(2) theta)`
`tan theta=sqrt(3)rArr theta=60^(@)` & `(g)/(2u^(2)cos^(2) theta)=2`
`rArr u=5/(2xx1/4)=sqrt(10)`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • KINEMATICS

    ALLEN|Exercise Comprehension#4|3 Videos

  • KINEMATICS

    ALLEN|Exercise Comprehension#5|6 Videos

  • KINEMATICS

    ALLEN|Exercise Comprehension#2|5 Videos

  • ERROR AND MEASUREMENT

    ALLEN|Exercise Part-2(Exercise-2)(B)|22 Videos

  • KINEMATICS (MOTION ALONG A STRAIGHT LINE AND MOTION IN A PLANE)

    ALLEN|Exercise BEGINNER S BOX-7|8 Videos

Similar Questions

Explore conceptually related problems

A particle is projected in x-y plane with y- axis along vertical, the point of projection being origin. The equation of projectile is y = sqrt(3) x - (gx^(2))/(2) . The angle of projectile is ……………..and initial velocity si ………………… .

The equation of trajectory of a projectile is y=10x-(5/9)x^2 if we assume g=10ms^(-3) then range of projectile (in metre) is

Knowledge Check

  • The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Range OA is :-

    A
    `sqrt(3)/2`
    B
    `sqrt(3)/4`
    C
    `sqrt(3)`
    D
    `3/8`

  • The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Time of flight of the projectile is :-

    A
    `sqrt(3/10)s`
    B
    `sqrt(10/3)s`
    C
    `1s`
    D
    `2s`

  • The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Maximum height H is :-

    A
    `8/3`
    B
    `3/8`
    C
    `sqrt(3)`
    D
    `2/sqrt(3)`

    • Similar Questions

    Explore conceptually related problems

    The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Radius of curvature of the path of the projectile at the topmost point P is :-

    The equation of a projectile is y=sqrt(3)x-(gx^(2))/(2) the angle of projection is:-

    The trajectory of a projectile in a vertical plane is y=ax-bx^(2) , where a and b are constants and x and y are respectively horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection form the horizontal are:

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

    The trajectory of a projectile in a vertical plane is y = ax - bx^2 , where a and b are constant and x and y are, respectively, horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projectile from the horizontal are.

    Home
    Profile