[" 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 "]

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

The equation of a projectile is y = ax - bx^(2) . Its horizontal range is

The equation of path of a particle projected obliquely from the ground under gravity is given by y=2x-x^(2) where 'x' and ' y' are in meter.The "x" and "y" axes are in horizontal and vertical directions respectively.The speed of projection is

For a particle,thrown from the ground obliquely, vec u=(5hat i+6hat j)ms^(-1) , vec a=(0hat i-10hat j)ms^(-2) where x and y axes are in horizontal and vertical directions respectively.The horizontal range of the particleis

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 trajectory of a projectile in a vertical plane is y=ax-bx^(2) , where a and b are constants and x and y are respectively horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection form the horizontal are:

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

[" 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]]

Equation of trajectory of a projectile is given by y = -x^(2) + 10x where x and y are in meters and x is along horizontal and y is vericall y upward and particle is projeted from origin. Then : (g = 10) m//s^(2)

A projectile is thrown with a velocity vec v _(0) = 3hati + 4 hatj ms ^(-1) where hat I and hatj are the unit vectors along the horizontal and vertical directions respectively. The speed of the projectile at the highest point of its motion in ms ^(-1) is