An object is projected with a velocity of `10 m//s` at an angle `45^(@)` with horizontal. The equation of trajectory followed by the projectile is `y = a x - beta x^(2)`, the ratio `alpha//beta` is

An object is projected with a velocitiy of 20m//s making an angle of 45^(@) with horizontal. The equation for trajectory is h=Ax-Bx^(2) where h is height x is horizontal distance. A and B are constants. The ratio A.B is x/0.1 . Find value of x.(g=10m//s^(2))

An object is projected with a velocity of 20 ms^(-1) making an angle of 45^(@) with horizontal. The equation for the trajectory is h = Ax -Bx^2 , where h is height, x is horizontal distance A and B are constants. The ratio A : B is (g = 10 ms^-2)

A body is projected with a velocity of 10 m//s at 45^(@) to the horizontal . The velocity of the projectile when it moves at 30^(@) to the horizontal is

Trajectory of a projectile is given by y = alphax + betax^2 . Then find (alpha + beta) .