The equations of motion of a projectile are given by x = 36t m and 2y=96t-9.8t2 m. The angle of projection is

Given: x = 36t and 2y = 96t - 9.8t2or y = 48t-4.9t2Let the initial velocity of projectile be u and angle of projection is θ. Then,Initial horizontal component of velocity,ux = u cosθ = (dxdt)t = 0   = 36  ---i or usinθ = 48-----(ii)Dividing (ii) by (i), we gettanθ = 4836  = 43sin θ = 45  or θ = sin-1(45)