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

The resultant velocity of a projectile at its highest point is sqrt(6/7) times that at half the maximum height. Show that the angle of projection with respect to the horizontal is 30^@ .

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.