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In many books, I see projectile motion caused by gravity when the object is in a state of free fall, so the velocity on the $y$ axis changes with respect to time and the velocity on $x$ axis is constant. But when the velocity on both the $x$ and $y$ axes change with respect to time, can the motion still be called projectile motion?

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An example of the type of motion you mention is the projectile motion with air resistance taken into account. In this case both vertical and horizontal components are functions of time. So this type of motion is of course possible and it is actually closer to the real behavior of a projectile.

Your question is about the name of this type of motion. The name is just a matter of convention. If you look up "projectile motion with air drag" you will see that it is pretty common to call this motion "projectile motion". So I would say that you can call it without danger of using an unusual terminology.

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Yes, it could be. The choice of axes is arbitrary.

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  • $\begingroup$ +1, but it might be helpful to explicitly point out a simple example — like where you just rotate the axes a little bit, so that the force of gravity has both $x$ and $y$ components. $\endgroup$ – Mike Sep 18 '18 at 2:46
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Yes and no. Projectile motion occurs when a single constant force is applied to a moving object. The force must be constant in magnitude as well as direction.

It is convenient to decompose the motion about two axes, one along the force and one perpendicular in order to simplify the solution, but you don't have to.

The resulting motion is a parabola, which we call projectile motion. The solution in vector form is:

$$ \boldsymbol{a} = \frac{1}{m} \boldsymbol{F} $$ $$ \boldsymbol{v} = \boldsymbol{v}_0 + \boldsymbol{a}\,t $$ $$ \boldsymbol{p} = \boldsymbol{p}_0 + \boldsymbol{v}_0 \,t + \frac{1}{2} \boldsymbol{a} t^2 $$

If along a particular direction the force (and hence the acceleration) is zero, then the position along that direction would be a result of constant motion.

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  • $\begingroup$ I take the OP to just mean that the velocity in the x-direction is changing rather than being constant (i.e. a horizontal force is acting on the object). It seems like you take it more to mean you take the basic projectile motion and then rotate your axes so that gravity has a component in both directions. Am I understanding your answer correctly? $\endgroup$ – Aaron Stevens Sep 17 '18 at 20:58
  • $\begingroup$ @AaronStevens - yes, because the two scenarios are equivalent. Put a horizontal constant force in the regular projectile and it acts like a single (combined) force acting in a skew dimension. This would be consistent with my answer which describe the necessary conditions for projectile motion. $\endgroup$ – ja72 Sep 17 '18 at 23:51
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Combining the answers of Chase Ryan Taylor and nasu:

Physicists usually consider coordinate systems a matter of choice, being unrelated to what actually happens, and only useful in describing and analysing it conveniently. Therefore the same "classic projectile motion" could be viewed from a coordinate system rotated with respect to the one usually used for this problem: for example the x axis could be paralel to the initial velocity of the projectile, or point in any arbitary direction. One could even use a 3D system (though this would usually be of little use (expect perhaps as a vector manipulation exercise))

However, interpreting the question the way you probably meant it (that the y axis is paralel with the local gravity vector, while x is perpendicular to it, being horizontal) the answer is still yes. If there is a horizontal force acting on the projectile so it is decelerating in this direction, it is still useful to label it "projectile motion", since this motion is also observable in projected (launched) objects (which move in some fluid)

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The necessary sufficient and only conditions for projectile motion to occour are-: 1)Velocity should remain constant along one of the axis 2)A constant acceleration must act along or opposite to the other axis. Hope that answers your question.

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    $\begingroup$ [citation needed]. $\endgroup$ – AccidentalFourierTransform Sep 17 '18 at 21:33
  • $\begingroup$ In case of air resistance the acceleration becomes a function of velocity and the equation of trajectory no longer is parabolic. What the textbooks mean by projectile motion is a parabolic trajectory.If there is acceleration (no matter whether it is constant or not) on both the axis instead of just being on 1 axis the trajectory of particle deviates from being parabolic. U can check it for yourself by taking constant accelerations a and b on both axis and writing the equation of trajectory. It won't be parabolic. $\endgroup$ – chemophilic Sep 17 '18 at 22:07
  • $\begingroup$ I have seen (French) textbooks speaking of projectile motion in presence of friction, and showing it is not parabolic. $\endgroup$ – Frédéric Grosshans Sep 20 '18 at 9:29
  • $\begingroup$ The english Wikipedia page on projectile motion agrees with your terminology, so I withdraw my (implicit) critics. It seems to be a country dependent terminology problem. $\endgroup$ – Frédéric Grosshans Sep 20 '18 at 9:33
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If you want to change the $x$ component of your velocity you would need to go in a different world. Anyway if you can do this we have three cases. The $x$ component becomes 0 before falling creating an increasing slope projectile it will fall of straightly after achieving zero velocity. Now it still has some deceleration on $x$ axis before falling to ground. Then it'll back off to it's throwing point same as a projectile. If it does fall before the $x$ velocity being zero then it'll be still a projectile. But while falling it'll have great slope. And through out it's journey it's slope will gradually increase. Now about equations if I think that gravity also acts in $x$ axis. $x= v\cos\theta t -gt^2/2 $ and $y= v\sin\theta t- gt^2/2$ $y-x= vt (\cos\theta -\sin\theta)$ so it's a straight line whose slope increasing or decreasing with the value $\theta$ with time.It also seems if $\theta = 45° $ then it'll get back to it's throwing point .

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  • $\begingroup$ The OP does not specify what force is changing the x-component of the velocity. It definitely is not gravity though. $\endgroup$ – Aaron Stevens Sep 17 '18 at 20:56
  • $\begingroup$ @AaronStevens: it could be gravity, if you do not neglect Earths curvature. Of course, neglecting friction and not neglecting planetary curvature implies quite literally you’re on another world ;-) $\endgroup$ – Frédéric Grosshans Sep 20 '18 at 9:36

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