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 $\mathrm{\theta }$ with the vertical at which the ball must be thrown, so that the ball returns to the boy’s hand is : (g = 10 m/s²)

$\begin{array}{l}{\mathrm{v}}_{\mathrm{y}}={\mathrm{v}}_0\cos \mathrm{\theta }=20\cos \mathrm{\theta }\\ {\mathrm{v}}_{\mathrm{x}}={\mathrm{v}}_0\sin \mathrm{\theta }=20\sin \mathrm{\theta }\end{array}$
Time of flight of the ball is :
$\mathrm{T}=\frac{2{\mathrm{v}}_{\mathrm{y}}}{\mathrm{g}}=\frac{40\cos \mathrm{\theta }}{10}=4\cos \mathrm{\theta }...\left(\mathrm{i}\right)$
In this time displacement of ball in horizontal direction should also be zero,
$\mathrm{i}.\mathrm{e}., 0={\mathrm{v}}_{\mathrm{x}}\mathrm{T}-\frac{1}{2}{\mathrm{a}}_{\mathrm{x}}{\mathrm{T}}^2$
This gives, $\mathrm{T}=\frac{2{\mathrm{v}}_{\mathrm{x}}}{{\mathrm{a}}_{\mathrm{x}}}=\frac{(20\sin \mathrm{\theta })2}{4}$
= 10 sin $\mathrm{\theta }$ .....(ii)
From eqn. (i) and eqn. (ii)
4 cos$\mathrm{\theta }$ = 10 sin $\mathrm{\theta }$
$\therefore \tan \mathrm{\theta }=\frac{4}{10}=0.4$