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

The total speed of a projectile at its greater height is sqrt(6/7) of its speed when it is at half of its greatest height. The angle of projection will be :-

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

A body is thrown at an angle theta_0 with the horizontal such that it attains a speed equal to (sqrt((2)/(3)) times the speed of projection when the body is at half of its maximum height. Find the angle theta_0 .

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

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))

When a projectile is at the highest point of its trajectory, the directions of its velocity and acceleration are

The speed of a projectile at its maximum height is sqrt3//2 times its initial speed. If the range of the projectile is n times the maximum height attained by it, n is equal to :

The launching speed of a certain projectile is five times the speed it has at its maximum height.Its angles of projection is