The horizontal range and the maximum height of a projectile are equal.If the speed of projection is `v_(0)`,then the speed at the highest point is `(2v_(0))/(sqrt(p))` The value of "p" is

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

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

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

If R and H are the horizontal range and maximum height attained by a projectile , than its speed of prjection is

The angle of projection at which the horizontal range and maximum height of projectile are equal is

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

Assertion: For projection angle tan^(-1)(4) , the horizontal range and the maximum height of a projectile are equal. Reason: The maximum range of projectile is directely proportional to square of velocity and inversely proportional to acceleration due to gravity.

The speed at the maximum height of a projectile is sqrt(3)/(2) times of its initial speed 'u' of projection Its range on the horizontal plane:-