a ball launched from ground level lands 2.85 s later on a level field 38 m away from the launch point. find the magnitude of the initial velocity vector and the angle it is above the horizontal. (ignore any effects due to air resistance.)

The magnitude of the velocity vector is 13.97m/s and the angle between at which the ball is launched is tan⁻¹(22.13).The horizontal displacement of the ball is 38m and the vertical displacement of the ball is zero.The time of the ball is 2.85 seconds.Using equation of motion, to find the vertical and horizontal component of the velocity,s = ut + 1/2at²s is the displacement,u is the initial velocity,a is the acceleration,t is the time.First for vertical component putting values,0 = u₁2.85 + 1/2(9.8)(2.5)²u₁ = 13.965m/s.Now, putting values for horizontal component,38 = u2.85 + 1/2(9.8)(2.85)²u₂ = 0.631 m/s.The velocity vector is 13.965m/si + 0.631m/sjNow, Magnitude = √(0.631)²+(13.965)²Magnitude = 13.97m/sHence, the magnitude of the velocity vector is 13.97m/s.The angle is given by,tanA = 13.965/0.631tanA = 22.13A = tan⁻¹(22.13)The angle at which it is launched is tan⁻¹(22.13).To know more about velocity vector, visit,https://brainly.com/question/626479#SPJ4