A boy throws a ball upwards with a speed $v_0=12m/s$. The wind imparts a horizontal acceleration of $0.4m/s^2$ to the left. If the ball must be thrown at an angle $\theta ={\tan }^{-1}\left(\frac{x}{5y}\right)$ so that it returns to the point of release, then find $(x+y)$

Step 1: Find time taken for horizontal motion.

Find time taken for horizontal motion.

Formula used: $t=\frac{2u}{a}$

Given, horizontal acceleration due to wind, $a_x=-0.4m/s^2$

And horizontal speed of the ball, $v_{x_i}=v_0\sin \theta$

Initial and final position of the ball, $x=0$

So, time taken by the ball for horizontal motion, $t=\frac{2v_0\sin \theta }{0.4}\dots \left(i\right)$

Step 2: Find time taken for vertical motion.

Formula used: $t=\frac{2u}{a}$

Given, vertical acceleration due to gravity, $a_y=-g$

And vertical speed of the ball, $v_{y_i}=v_0\cos \theta$

Initial and final position of the ball, $y=0$

So, time taken by the ball for vertical motion,
$t=\frac{2v_0\cos \theta }{g}\dots \left(ii\right)$

Equating $\left(i\right)$ and $\left(ii\right)$, we get $\tan \theta =\frac{1}{25}$

$\Rightarrow \theta ={\tan }^{-1}\left(\frac{1}{25}\right)$

Compare this with $\theta ={\tan }^{-1}\left(\frac{x}{5y}\right)$

So, $(x+y)=6$

Final answer: 6