A boy throws a ball upward with velocity vo = 20 m/s. The wind imparts a horizontal acceleration of 4 m/s2 to the left. The angle θ with the vertical at which the ball must be thrown so that the ball returns to the boy's hand is: (g = 10 m/s2)

vy = v0cosθ = 20 cosθvx = vosinθ = 20 sinθTime of flight of the ball is :T = 2vyg = 40 cosθ10 = 4 cosθ ---(i)In this time displacement of ball in horizontal direction should also be zero,i.e., 0 = vxT-12axT2This given, T = 2vxax = (20 sinθ)24                     = 10 sinθ --(ii)From equations (i) and (ii),4 cosθ = 10 sinθtan θ = 410 = 0.4