A boy throws a ball upwards with velocity u=15ms-1. The wind imparts a horizontal acceleration of  3ms-1 to the left  as shown. The angle  θ at which the ball must be thrown so that the ball returns to the boy’s hand is (use  g  = 10ms-1)

here uy=ucosθ=15cosθ   ux=usinθ=15sinθ time of flight =2uyg=2×15cosθ10=3cosθ---(1) in this time displacement of ball horizontally=0 sx=uxt+12axt2 0=uxt-12axt2    negative since wind acceleration is to left t=2uxax=2×15sinθ3=10sinθ---(2) equating (1) and (2) 3cosθ = 10sinθ tanθ=0.3 θ=tan-1(0.3)