# 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.

• I'll just say that velocities in x and y direction (orthogonal) are independent. look deeper, you're misinterpreting something here as far as i can see. Oct 29, 2013 at 21:22
• Oh! When the angle that one is firing from is 0 so that for example the barrel of a gun is parallel to the ground, the time it takes for the projectile to hit the ground will then be the same for all velocities.
– Reds
Oct 29, 2013 at 21:31
• But I still dont know why there should be any difference if you tilt the barrel
– Reds
Oct 29, 2013 at 21:33
• See also this MythBusters episode youtube.com/watch?v=D9wQVIEdKh8 Oct 29, 2013 at 21:41
• @Reds I think that if you consider friction the velocity would be important. Oct 29, 2013 at 22:29

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.

Maybe you should try this for a (thought)experiment:
Imagine you're on an "invisible" train that's initially motionless. You drop a ball in the $y$ direction towards the ground and measure the time it takes to get to ground. Then the train starts to move. After a while, when the train is going with a constant velocity of $v_0$, you drop the same ball again, in the same direction towards the ground. I imagine you wont feel anything weird in the motion of the ball while doing this. Now imagine a friend watching outside the "invisible" train. When the invisible train is moving, he would see at some point the ball is thrown horizontally with a velocity of $v_0$. You and your friend are seeing the same ball hit the ground. If the time it takes increases by increasing $v_0$, then either you and your friend will not agree on the time you measured, or assuming your friend is correct, you would feel that the ball is taking longer than usual to reach the ground, and it would get worse if the train started to go faster.

Because you can separate the vector space (assumption) of R x R into two separate vector spaces, each spanning the real line R with independent basis. When you are in the vector space R then you can't reach all elements in R x R. You can do that if you make a superposition of each of these R spaces into R x R and then you can reach all elements in this space with the help of the spaces R. Generally speaking.