# Horizontal velocity of a projectile [closed]

I was trying to prove that horizontal velocity of a projectile remains constant mathematically, but can't fully come close to proving it.

Is it possible to do it, or do we have to live with this assumption which is proven correct only experimentally.

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## closed as too localized by Qmechanic♦, Emilio Pisanty, Waffle's Crazy Peanut, Sklivvz♦, twistor59Jan 4 '13 at 19:53

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Could you post your derivation here so we can see how far you've gotten? – Kitchi Nov 14 '12 at 10:46
Hint: What is the sum of the forces acting in the horizontal direction? – ja72 Nov 16 '12 at 21:44

## 2 Answers

You certainly cannot prove it mathematically because it is called the Netwon's first law, which is proved by experiment.

It states that an object with no external force will move with constant velocity. Since $x$ and $y$ component are independent, so the horizontal velocity is certainly constant (when there is no friction).

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Then what is the external force in its y component, its velocity should be constant too ? I haven't seen any satisfactory answer to this till now, people keep repeating what they read in the books. – pokrate Dec 16 '12 at 4:22
The external force is the gravity $g$, which only act on y component. Note that the velocity is not constant. Also, the vertical velocity is not constant, only the horizontal velocity is constant. It is very important for you to know that x and y component can be treated independently. As there is no force in the horizontal direction, you know that the horizontal velocity is constant by Newton first law – hwlau Dec 16 '12 at 4:29

I'm guessing you're looking at a projectile or something similar where the force is acting only in a vertical direction. If so the proof is very simple. The infinitesimal change in velocity is related to the acceleration by:

$$\vec{dv} = \vec{a} dt$$

If we write the vectors out as horizontal and vertical components we get two equations:

$$\vec{dv_x} = \vec{a_x} dt$$

$$\vec{dv_y} = \vec{a_y} dt$$

And $\vec{a_x} = \vec{F_x}/m$ where $\vec{F_x}$ is the horizontal component of the force and $m$ is the projectile mass. Since the gravitational force is vertical $\vec{F_x} = 0$ and therefore $\vec{a_x} = 0$ and therefore $\vec{dv_x} = 0$ i.e. the change in the horizontal velocity is zero.

But this is such a simple argument I wonder if I have misinterpreted your question. If so add a comment or edit your question and I'll try to respond.

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