newtonian mechanics - Does $y$-motion really have nothing to do with $x$-motion?

I am watching a Physics 1 for physical science majors on coursera.com, and in one of the concept tests there is a question that goes like this; "A bullet is fired horizontally from a rifle on the Moon (where there is no air). The initial speed of the bullet when it leaves the gun barrel is $V_{0}$. Assume that the ground is perfectly level (and endless)."

And then there's 3 statements, on of which is; III) The time it takes for the bullet to hit the ground increases as $V_0$ is increased.

Apparently this statement is false.

The lecturer shows a simulator here where a cannon shoots a ball in a air free environment. And when the lecturer checks the time that it took for the ball to hit the ground after several shots with different initial velocities, the time is the same. Not really understanding this, I tried it for myself because I thought; "Surely, the time it takes for the ball to hit the ground is increased as the initial velocity is increased, at least if I were to increase the initial velocity by a large amount." So I did. And sure enough the time for the ball to hit the ground did increase.

How is that, and isn't it inconsistent with what the lecturer says about $x$-motion having nothing to do with $y$-motion?'

The lecturer has also showed the equation; $$x = x_0+ v_{0x}t+\frac{1}{2}a_xt^2$$ $$ a = 0 \space \vec{} \space x=x_0+v_0\cos(\theta) t$$ $$a = -g \space \vec{} \space y = y_0+v_0\sin\theta-\frac{1}{2}gt^2$$ Showing that you can treat $x$- and $y$-motion independently.

Answer

The time it takes is independent of the initial velocity if and only if the barrel is horizontal to the ground. If the barrel is horizontal, then the initial velocity will be solely in the $x$-direction and the only variable affecting the $y$-direction will be the acceleration due to gravity.

However, if the barrel is at an angle, then the initial velocity will have a vertical component. If $\theta$ is the angle from the ground, we can express the initial velocity in the $y$-direction as:

$v_{0, y} = v_0\sin(\theta)$

We can also express the height $y$ above the ground using the free-fall equation:

$y=v_{0,y}t - \frac{1}{2}gt^2=v_0\sin(\theta)t - \frac{1}{2}gt^2$

Notice that an increase in $v_0$ results in an increase in the amount of time it takes for the object to reach the ground. However, if $\theta=0$, then $v_0\sin(\theta)t$ will always equal $0$ because $\sin(0) = 0$. If $v_0\sin(\theta)t=0$, we can cancel it out of the equation and get $y = -\frac{1}{2}gt^2$, which is independent of the initial velocity.