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

The equation of a trajectory of a projectile is y=8x-4x^(2) The maximum height of the projectile is (A) 8m (B) 6m (C) 2m (D) 4m

The equation of a projectile is y = ax - bx^(2) . Its horizontal range 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 projectile is y = 16 x - (5)/(4)x^(2) find the horizontal range

The equation of a projectile is y = sqrt(3)x - ((gx^2)/2) the horizontal range is

The equation of projectile is y=16x-(x^(2))/(4) the horizontal range is:-

The equation of a projectile is y=sqrt(3)x-(gx^(2))/(2) the angle of projection is:-

The parabolic path of a projectile is represented by y=x/sqrt3-x^(2)/60 in MKS units: Its angle of projection is (g=10ms^(-2))

The equation of motion of a projectile is y = 12 x - (3)/(4) x^2 . The horizontal component of velocity is 3 ms^-1 . What is the range of the projectile ?