The equation of path of a particle projected obliquely from the ground under gravity is given by `y=2x-x^(2)` where 'x' and ' y' are in meter.The "x" and "y" axes are in horizontal and vertical directions respectively.The speed of projection is

[" The equation of trajectory of a projectile is "],[y=ax-bx^(2)" where "x" and "y" axes are along "],[" horizontal and vertical directions "],[" respectively.The range of the projectile is "]

For a particle,thrown from the ground obliquely, vec u=(5hat i+6hat j)ms^(-1) , vec a=(0hat i-10hat j)ms^(-2) where x and y axes are in horizontal and vertical directions respectively.The horizontal range of the particleis

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)

Trajectory of particle in a projectile motion is given as: y=x-(x^(2)//80) (x and y are in meters). Take g=10m//s^(2) .

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 position coordinates of a projectile projected from ground on a certain planet (with an atmosphere) are given by y = (4t – 2t^(2)) m and x = (3t) metre, where t is in second and point of projection is taken as origin. The angle of projection of projectile with vertical is -

A body is projected horizontally from a point above the ground. The motion of the body is described by the equations x = 2 t and y= 5 t^2 where x and y are the horizontal and vertical displacements (in m) respectively at time t. What is the magnitude of the velocity of the body 0.2 second after it is projected?

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 vertical and horizontal displacements of a projectile thrown from a point on the ground at time t are given by y=6t-5t^(2) and x=8t respectively,where x and y are in meters and t is in seconds.The time of flight of the projectile in seconds is ?