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

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

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 :-

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 :-

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

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

The equation of trajectory ofa projectile is y = 10x-5/9x^2 If we assume g = 10 ms^(-2) , the range of projectile (in metre) 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.

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.

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 is ………………… .

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 is ………………… .

Home
Profile