A body throws a ball upwards with velocity ${\upsilon }_0=20m{s}^{-1}$ . The wind imparts a horizontal acceleration of  $4m{s}^{-2}$ to the left. If the angle  at which the ball must be thrown so that the ball returns to the boy’s hand is ${\tan }^{-1}\left(\frac{x}{10}\right)$ then $x$ is  $(g=10m{s}^{-2})$
                                      

If T be the time of flight of the ball, then 
$T=\frac{2v_y}{g}=\frac{2\left(V_0\cos \theta \right)}{g}$

$\Rightarrow T=\frac{2\left(20\cos \theta \right)}{10}=4\cos \theta$                …………… (1)
For the ball to return to the boy’s hand, we have  $x=0$
       

                                   
$\Rightarrow 0=v_{x}T+\frac{1}{2}{a}_x{T}^2$
$\Rightarrow T=10\sin \theta$                                        …………… (2)
From (1) and (2), we get 
$\Rightarrow 0=\left(v_0\sin \theta \right)T+\frac{1}{2}(-4)T^2$

$\Rightarrow 0=20\sin \theta -2T$
 
$4\mathrm{cos\theta }=10\mathrm{sin\theta }$

$\Rightarrow \theta ={\tan }^{-1}\left(\frac{4}{10}\right)$