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

Given : x=36t, 2y=96t−9.8t2 or  y=48t−4.9t2Let the initial velocity of projectile be u and angle of projection is θ. Then,Initial horizontal component of velocity, ux=ucos ⁡θ=dxdtt=0 =36 or u cos ⁡θ=36           ....(i)Initial vertical component of velocity, uy=usin⁡θ=dydtt=0 =48 or u sin ⁡θ=48              …..(ii)Dividing (ii) by (i), we get tan ⁡θ=4836=43; ∴ sin⁡θ=45  or  θ=sin−1⁡45