A boy throws a ball upward with velocity $v_o = 20 m/s.$ The wind imparts a horizontal acceleration of 4 $m/s^2$ to the left. The angle $\theta$ with the vertical at which the ball must be thrown so that the ball returns to the boy's hand is: (g = 10 $m/s^2$)

$v_y = v_0\cos \theta = 20 \cos \theta$

$v_x = v_o\sin \theta = 20 \sin \theta$

Time of flight of the ball is :

$T = \frac{2v_y}{g} = \frac{40 \cos \theta }{10} = 4 \cos \theta ---\left(i\right)$

In this time displacement of ball in horizontal direction should also be zero,

$i.e., 0 = v_{x}T-\frac{1}{2}{a}_x{T}^2$

This given, T = $\frac{2v_x}{a_x} = \frac{(20 \sin \theta )2}{4}$

                     = $10 \sin \theta --\left(ii\right)$

From equations (i) and (ii),

$4 \cos \theta = 10 \sin \theta$

$\tan \theta = \frac{4}{10} = 0.4$