If the range of a projectile is `x` times its maximum height attained and also its speed at maximum height is 1/2 times its speed of projection, then the value of `x` is

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

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

The speed of a projectile when it is at its greatest height is sqrt(2//5) times its speed at half the maximum height. The angle of projection is

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

For which angle of projection the horizontal range is 5 times the maximum height attrained ?

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

Two balls are thrown with the same speed from a point O at the same time so that their horizontal ranges are same. If the difference of the maximum height attained by them is equal to half of the sum of the maximum heights, then the angles of projection for the balls are

Find the angle of projection for a projectile if its speed at maximum height is sqrt(5//9) times its speed at one-third of maximum height (g=10ms^(-2))