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While studying motion in 2D, I came to know that when an object, projected with some initial velocity, making angle $0°$ with the horizontal, will have zero range.

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Its a bit confusing. The object has a velocity component along the $x$-axis (horizontal). Hence, since acceleration along $x$-axis is 0, I think it will travel distance according to the equation: $$U=\frac{X}{T}$$ where $U$ is the initial velocity, $X$ is the distance travelled by the object and $T$ is the total time.

Is my reasoning correct? I just want to how exactly the range of a projectile is defined.

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  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented May 28 at 13:10
  • $\begingroup$ The range is the distance between the two points where the trajectory crosses the horizontal axis. But if the body is projected at 0° it never leaves the horizontal axis. $\endgroup$
    – PM 2Ring
    Commented May 28 at 13:13
  • $\begingroup$ Thanks for your answer.It seems like i misunderstood the actual meaning of range $\endgroup$
    – Ishaan
    Commented May 28 at 13:47
  • $\begingroup$ Skipping derivation, rage covered by projectile $$ x = \frac {v_0^2 \sin(2\alpha)}{g} $$ where $\alpha$ is angle between initial velocity vector and horizontal. Note, your initial speed $u$ is projection into x coordinate axis, while usually projectile motion is defined using initial velocity vector (not projection of it into $x$ or $y$ axis). $\endgroup$ Commented May 29 at 6:41
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    $\begingroup$ en.wikipedia.org/wiki/Range_of_a_projectile $\endgroup$
    – PM 2Ring
    Commented May 29 at 9:25

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A projectile is typically defined as an object given some initial impulse, which travels through space under only the force of gravity and possibly drag or thrust. Once an object hits the ground, it's not a projectile anymore. The range of a projectile typically refers to the horizontal distance over which the projectile travels while in the air. A projectile fired at $0$ degrees from a height of $0$ m strikes the ground immediately upon being fired, and never leaves the ground - it travels exactly zero meters in the air.

In reality the object will probably strike the ground and skid or bounce over some distance, but most physics problems don't deal with the complicated interaction between a projectile and the ground. Even if the object's behavior after impact is defined, projectile range will generally only refer to the behavior between launch and impact.

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  • $\begingroup$ The range of a projectile typically refers to the horizontal distance over which the projectile travels ''while in the air''.So for range to be defined the object must be in air.I think now i got the answer to my question.Thanks! $\endgroup$
    – Ishaan
    Commented May 28 at 15:44
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    $\begingroup$ @Ishaan The range is still defined even if the object never leaves the ground, it's just a range of 0. A better example of undefined range might be for an object launched at escape velocity - its range is infinite and has no defined value. $\endgroup$ Commented May 28 at 16:00
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With an angle of zero degrees, I assume that you mean that you are dealing with horizontal motion - that being motion only along the $x$ axis - with no vertical movement up or down. Maybe this is a cart on wheels, or the like, which you have pushed sideways. If there is no friction or other horizontal forces involved, then I agree with you that there will be no acceleration influencing this motion. That means that a kinematic motion equation like $x=x_0+ut+\frac12at^2$ would reduce to $x=ut$ (we place our coordinate system such that $x_0=0$), and then you have your formula $u=x/t$. I agree fully with all this.

The question now is, what does the term range refer to?

In mathematics, this term range refers to the set of all function values that are possible to get from a function. Meaning, "all possible outputs", so to say, which also sometimes is refered to as the image of a function.

I am guessing from your text that maybe we are supposed to imagine an elevation function (a $y$ function) that outputs $y$ values. Then the range would be all $y$ values that are possible to reach. If your motion is purely along the $x$ axis, as discussed above, then there is no change in the $y$ coordinate. If the motion starts at the origin where $y=0$, then the range would indeed be $0$ (since this is the only $y$ value that is possible to achieve).

This is guess-work, though, since it all depends on in what context this word range is used. So, have a check in your textbook (or ask the teacher who said this) for where this term is defined more accurately, or simply check for which exact function they are refering to when using the term range.

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  • $\begingroup$ Thank you for your answer.By range i refer to the horizontal distance the projectile travels from the time it is launched to the time it comes back down to the same height at which it is launched(Definition taken from internet) $\endgroup$
    – Ishaan
    Commented May 28 at 15:42
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    $\begingroup$ @Ishaan I find that definition a little restrictive - the range won't necessarily require that the projectile finish at the same height it was launched from (although it is common to find the range on flat ground). A projectile launched from the top of a mountain, for example, will have greater range than one launched from flat ground, due to the additional vertical distance/flight time. Range is usually the horizontal distance between the start and end of the trajectory, and there's nothing that requires the trajectory to end when the projectile comes back down to the launch height. $\endgroup$ Commented May 28 at 18:17
  • $\begingroup$ @Ishaan Well, if that is the definition of range, then the range must obviously be zero in your setup, since there is no movement along the y axis. Now I do not understand your question, to be honest - I thought you were asking to how you should understand this term range, but here I see that you already have a clear definition of it? $\endgroup$
    – Steeven
    Commented May 29 at 5:22

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