Trajectory of a particle in a projectile motion is given as `y=x-(x^(2))/80`. Here `x` and `y` are in metre. For this projectile motion the following with `g=10m//s^(2)`.
Text Solution
Verified by Experts
The correct Answer is:
A, B, C, D
Topper's Solved these Questions
KINEMATICS
DC PANDEY ENGLISH|Exercise INTEGER_TYPE|10 Videos
KINEMATICS
DC PANDEY ENGLISH|Exercise COMPREHENSION_TYPE|13 Videos
GRAVITATION
DC PANDEY ENGLISH|Exercise (C) Chapter Exercises|45 Videos
KINEMATICS 1
DC PANDEY ENGLISH|Exercise INTEGER_TYPE|15 Videos
Similar Questions
Explore conceptually related problems
Trajectory of particle in a projectile motion is given as y=x- x^(2)/80 . Here, x and y are in meters. For this projectile motion, match the following with g=10 m//s^(2) . {:(,"Column-I",,"Column-II"),((A),"Angle of projection (in degrees)",(P),20),((B),"Angle of velocity with horizontal after 4s (in degrees)",(Q),80),((C),"Maximum height (in meters)",(R),45),((D),"Horizontal range (in meters)",(S),30),(,,(T),60):}
Trajectory of particle in a projectile motion is given as y = x - (x^(2))/(80) . Here x and y are in metre. For this projectile motion match the following with g = 10 ms^(-2) .
The potential energy (in joules ) function of a particle in a region of space is given as: U=(2x^(2)+3y^(2)+2x) Here x,y and z are in metres. Find the maginitude of x compenent of force ( in newton) acting on the particle at point P ( 1m, 2m, 3m).
The displacement x of a particle in a straight line motion is given by x=1-t-t^(2) . The correct representation of the motion 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 :-
If y=x-x^2 is the path of a projectile, then which of following is incorrect ( 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)] Range OA is :-
Equation of trajectory of a projectile is given by y = -x^(2) + 10x where x and y are in meters and x is along horizontal and y is vericall y upward and particle is projeted from origin. Then : (g = 10) m//s^(2)
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 :-
Acceleration of a particle under projectile motion at the highest point of its trajectory is :