The equation of a trajectory of a projectile is `y=8x-4x^(2)`. The maximum height of the projectile is ?

At the top of the trajectory of projectile the

The equation of the path of the projectiles is y=0.5x-0.04x^(2) .The initial speed of the projectile is? (g=10ms^(-2))

The equation of trajectory of a projectile is y=10x-(5/9)x^2 if we assume g=10ms^(-3) then range of projectile (in metre) is

The equation of trajectory of an oblique projectile y = sqrt(3) x - (g x^(2))/(2) . The angle of projection is

[" The equation of trajectory of a projectile is "],[y=ax-bx^(2)" where "x" and "y" axes are along "],[" horizontal and vertical directions "],[" respectively.The range of the projectile is "]

Trajectory Of Horizontal Projectile

[" The equation of a trajectory of a "],[" projectile is "],[[y=3x-0.02x^(2)," where "'x'" and "'y'],[" are in meters,range of projectile is "],[" (A) "150m],[" (B) "300m],[" (C) "450m],[" (D) "600m]]

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