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

During a projectile motion if the maximum height equal the horizontal range, then the angle of projection with the horizontal is :

If R and H represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal 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 ratio of the speed of a projectile at the point of projection to the speed at the top of its trajectory is x. The angle of projection with the horizontal is

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

A projectile is projected from ground such that the maximum height attained by, it is equal to half the horizontal range.The angle of projection with horizontal would be

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