The potential energy of a projectile at maximum height is 3/4 times kinetic energy of projection.Its angle of projection is (A) `30^@`(B)`45^@`(C) `60^@` (D) None of these

The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is .

The potential energy of a projectile at its maximum height is equal to its kinetic energy there. If the velocity of projection is 20 m//s^(-1) ,its time of flight is (g=10 m//s^(2))

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

The horizontal range of a projectile is 4 sqrt(3) times its maximum height. Its angle of projection will be

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

The kinetic energy of a projectile at the highest point is half of the initial kinetic energy. 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 horizontal range is equal to two times maximum height of a projectile. The angle of projection ofthe projectile is: