Home
Class 11
PHYSICS
The parabolic path of a projectile is re...

The parabolic path of a projectile is represented by `y=x/sqrt3-x^(2)/60` in `MKS` units: Its angle of projection is `(g=10ms^(-2))`

A

`30^(@)`

B

`45^(@)`

C

`60^(@)`

D

`90^(@)`

Text Solution

Verified by Experts

The correct Answer is:
B
`y=Ax -Bx^(2), theta =tan^(-1)(A)`
Promotional Banner

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)] Angle of projection theta is :-

The equation of the path of the projectiles is y=0.5x-0.04x^(2) .The initial speed of the projectile is? (g=10ms^(-2))

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

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

The horizontal range of a projectile is 2 sqrt(3) times its maximum height. Find the angle of projection.

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

Equation of trajector of ground to ground projectile is y=2x-9x^(2) . Then the angle of projection with horizontal and speed of projection is : (g=10m//s^(2))

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)] Range OA is :-