The time of flight of a projectile is related to its horizontal range by the equation `gT^(2)=2R`.The angle of projection is?

The equation of a projectile is y = ax - bx^(2) . Its horizontal range is

If the time of flight of a bullet over a horizontal range R is T, then the angle of projection with horizontal is -

Prove that the time of flight T and the horizontal range R of a projectile are connected by the equation: gT^2 =2 R tan theta , where theta is the angle of projection.

For a given angle of projection, if the initial velocity of a projectile is doubled, then its horizontal range will become ____ times the original horizontal range. ____.

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

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

The time of flight, for a projectile, along two different paths to get a given range R, are in ratio 2 : 1 . Find the ratio of this range R to the maximum possible range for the projectile assuming the projection speed to be same in all cases.

The range of a projectile is R when the angle of projection is 40^(@) . For the same velocity of projection and range, the other possible angle of projection is