The path of a projectile is given by the equation `y = ax – bx^(2)`, where a and b are constants and x and y are respectively horizontal and vertical distances of projectile from the point of projection. The maximum height attained by the projectile and the angle of projection are respectively.

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 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 horizontal range is four times the maximum height attained by a projectile. The angle of projection is

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

If the horizontal range of a projectile be a and the maximum height attained by it is b, then prove that the velocity of projection is

The potential energy of a projectile at maximum height is 3/4 times kinetic energy of projection.Its angle of projection

Show that the horizontal range of a projectile is same for two angles of projection . Is the sum of maximum heights for these two angles dependent on the angle of projectile ?

The horizontal range is equal to two times maximum height of a projectile. The angle of projection ofthe projectile is:

If R and H represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal is