Cody's "Angry LADIES" ! ....not birds....

The displacement along Y axis is $-200m$ . Initial velocity along Y axis is 0.

$S_y = -200 = ut - \frac{gt^2}{2}$ ... here $u=0$ and $g=10$

$t^2 = 40 \implies t =\sqrt{40} seconds$

Acceleration along X axis is 0, and initial velocity is 10m/s as given.

Observe that if the walls were not there, the chocolate would have gone $S_x = ut + \frac{at^2}{2} = 10t = 10\sqrt{40} m$

Here in this case, according to given conditions of collision, note that if you unfold that path, it is same as the path without walls, hence here , the horizontally Distance covered (not displacement) is same as $10\sqrt{40}$

Thus the chocolate moved horizontally a distance of $10\sqrt{40} m$ during the motion and as the distance between walls is $10m$ , the number of times it's path is changed, i.e. it hits the wall is

$\lfloor \frac{10\sqrt{40}}{10} \rfloor = \lfloor \sqrt{40} \rfloor = 6$

Thus it hits the walls $\boxed{6}$ times and , before it could hit the wall 7th time, it reaches the ground.

its x-axis velocity is constant,so it will collide once per second.The time needed to fall in the ground is t=sqrt(2h/g)=sqrt(40),6<t<7.So in time t it will strike the walls 6 times.