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 equations of motion of a projectile are given by x=36tm and 2y=96t-9.8t^(2)m .The angle of projection is

The equations of motion of a projectile are given by x=36tm and 2y =96t-9.8t^(2)m . The angle of projection 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]]

The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Time of flight of the projectile is :-

The maximum horizontal range of a projectile is 400 m . The maximum value of height attained by it will be

The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Maximum height H is :-

" Maximum height attained by an oblique projectile "(h_(m))

The equation of trajectory of a projectile thrown from a point on the ground is y=(x-x^(2)/40) m. If g=10ms^(-2) The maximum height reached is

The line y=mx+1 is a tangent to the parabola y^2 = 4x if (A) m=1 (B) m=2 (C) m=4 (D) m=3

The time of flight of a projectile is 10 s and range is 500m. Maximum height attained by it is [ g=10 m//s^(2) ]