A boy throws a ball upwards with velocity v0 = 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=v0cos⁡θ=20cos⁡θvx=v0sin⁡θ=20sin⁡θTime of flight of the ball is :T=2vyg=40cos⁡θ10=4cos⁡θ...(i)In this time displacement of ball in horizontal direction should also be zero,i.e.,  0=vxT−12axT2This gives, T=2vxax=(20sin⁡θ)24= 10 sin θ .....(ii)From eqn. (i) and eqn. (ii)4 cosθ = 10 sin θ∴tan⁡θ=410=0.4