# The effect of gravity on a speeding bullet

Discussion in 'Physics & Math' started by Magical Realist, Jan 31, 2017.

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Correct!

3. ### SarkusHippomonstrosesquippedalo phobeValued Senior Member

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No, simply distinguishing between the various vectors of the velocity (vertical, horizontal, net).
In the current scenario of the bullet, yes, that has not been disagreed with.
And if there is no horizontal velocity then the net vertical velocity is the net terminal velocity.
This is simple pythagoras' theorem at work.

5. ### Magical RealistValued Senior Member

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Somehow I'm back now. Sorry to ruin your day exchemist. So are you saying the bullet would continue to accelerate in a vacuum? What force is causing it to accelerate? Doesn't acceleration always involve a force?

Last edited: Feb 10, 2017

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Gravity.

8. ### sculptorValued Senior Member

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Seems accurate.
thanx

9. ### Magical RealistValued Senior Member

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Is gravity a force?

10. ### RajeshTrivediValued Senior Member

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Thanks.
You are right depending on the object (shape size volume mass density etc, and of course airflow and altitude) it may not become zero. And thats what I am saying that as long as this horizontal component is non zero the object cannot achieve terminal velocity. Few posters are stuck here with high school orthogonal force and velocity vector calculations without understanding the vagaries of drag force. They must attempt to appreciate that resultant drag force cannot be calculated by the velocity components. For example for drag force,

F(resultant) may not be equal to Sqrt(Fx^2 + Fy^2), where Fx and Fy are calculated from the Vx and Vy. Thats the complexity of drag dependence on velocity and other factors. Even the drag coefficient is speed dependent. let the V^2 component in drag force for bullet etc, not confuse you.

11. ### RajeshTrivediValued Senior Member

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Stay ignorant, no problem.
but more important stay blessed..

12. ### originTrump is the best argument against a democracy.Valued Senior Member

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Yes, gravity acts like a force, that is exactly the way Newton modeled it. If you released a mass very far from Earth it would accelerate until it hit the earth. The speed of the object when it hits the earth would be about 25,000 mph.

Edit to add - of course that is neglecting friction!

13. ### sculptorValued Senior Member

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Thanks for the edit, it cured my
HUH???

14. ### dumbest man on earthReal Eyes Realize Real LiesValued Senior Member

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No thanks needed, RajeshTrivedi , but still appreciated. And, you are more than welcome
No confusion, either.

15. ### SarkusHippomonstrosesquippedalo phobeValued Senior Member

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Um, who said that the sum would be the same??? Are you sure you even understand what people are writing? The forces on an object travelling 200i + 800j would be the square-root of the sum of the forces-squared of 200i and 800j. Simple Pythagoras theorem gives you that.
Yes, the issue is complicated in ensuring that you use the correct variables when calculating the horizontal and vertical components, due to object shape etc. Airflow would come into it if not assumed homogenous (I.e. at any given point the air behaves the same in one direction as another). It would likely be easier to directly calculate the drag in the direction of flow at any given moment, though, but just as with any other force, it can be split into horizontal and vertical components.

Further, you are introducing aerodynamic drag, where motion in one direction affects forces in the other, yet you seem blind to the realities of it by asserting that horizontal motion must be zero before terminal velocity can be reached. Which do you want to assert: that you're discussing aerodynamic drag or that horizontal motion must be zero?

16. ### RajeshTrivediValued Senior Member

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Ummm! still you are out of your depth.

1. Your assertion that the bullet will achieve the terminal velocity even with non zero horizontal speed is incorrect. Please note that in case of drag scenario the terminal speed can only be achieved if the net acceleration on the object is zero. Your claim that let the force be there in horizontal direction, vertical direction will still achieve terminal stable speed, is incorrect.

2. Your fixation with Pythagoras is misplaced here. If the falling bullet has a velocity vector 200i + 800 j (Consider the frame accordingly), then the resultant speed is 824 meter / seconds. Now drag force F(824) is not equal to Mag[F(200)i + F(800)j]. If you appreciate this, then you will agree to #1 above.

17. ### RajeshTrivediValued Senior Member

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If you are challenging my knowledge then be specific or be ready for proper discussion. Otherwise you are just trolling and flaming.

18. ### SarkusHippomonstrosesquippedalo phobeValued Senior Member

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I am not necessarily asserting that this is true of the bullet, nor have I ever done, but it could be true of certain objects as it will depend upon the aerodynamics of that object.
It is quite possible that an object can travel at terminal velocity that has a non-zero horizontal speed. All that needs to happen is that as the object falls through the air the aerodynamics produce a horizontal force as well as vertical drag. If the net of the created forces balances the force due to gravity then you end up with net zero force in any direction but with a forward and downward velocity.
Do you not comprehend that the horizontal speed does not need to be zero for there to be a net zero force in the horizontal direction? After all, any aircraft travelling at constant horizontal speed has a net zero force: the drag is balanced by the forward thrust of the engines.
Now, replace that thrust by the aerodynamic force created as the object falls downward. Surely you can imagine that, given that aircraft have wings that are generally used to convert forward motion into lift?
You misunderstand the reference to Pythagoras; it merely helps identify the vector of the net force. As long as that vector balances the force due to gravity, as well as zero net horizontal force, the object will be at terminal velocity, irrespective of actual direction of travel.

So yes, it can be the same, depending upon the aerodynamics of the object. For a round ball, no, it won't be, but then a round ball is not aerodynamic, and a bullet just as unlikely. For an aerofoil that presents very different cross sections in the horizontal and vertical, and has the appropriate aerodynamic properties, yes it can be the same.
Bear in mind that while the drag force is proportional to the square of the velocity it is also proportional to the cross-sectional area as well as the coefficient of drag. And depending upon the shape of the object, this coefficient can lead to what you seem to refuse to accept.

Do you accept that for an object to have achieved terminal velocity there simply need be zero net force on the object?
Do you accept that aerodynamic objects can travel through the air in one direction and create a force in another?

Okay, so say a mass of 10kg is falling through the air vertically. It will achieve terminal velocity when drag matches the gravitational force of 100N, agreed?
Right, so say the mass is also travelling through the air horizontally but is unaerodynamic so that the drag it creates slows down the horizontal motion. This is, we agree, the case of the bullet.
Now imagine that the mass is aerodynamic, and has achieved terminal velocity according to the notion above... As it falls at the angle it is, it is producing lift of L perpendicular to the direction of travel, and drag D in the opposite direction of travel, right? So, using Pythagoras' theorem, that you seem so loathe to consider, can you not see how the net of the L and D could balance each other horizontally yet also balance the 100N due to gravity?

Using a simplistic example, let's say the angle of descent is 45-degrees, and let's say it generates drag of 70.71N and Lift of 70.71N as well.

Now, while it should be self-evident that the horizontal components of the Lift and Drag cancel each other out, using Pythagoras it can be established that the vertical component of the Lift and Drag is 100N in the opposite direction to the 100N of force due to gravity.
Hence you end up with an object with zero net force (I.e. Terminal velocity) yet with non-zero horizontal speed.

Yes, these numbers are simplistic but they are just to demonstrate that it is possible.

So what of this do you wish to dispute?

19. ### RajeshTrivediValued Senior Member

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Good, so we agree on this.

Did I say anything different? I said that net acceleration to be zero. Good, here also we agree.

I am not disputing anything. I just stated the physics around a bullet shot from a flying aircraft. You have come around in this post only, otherwise you were disputing my stand.

20. ### SarkusHippomonstrosesquippedalo phobeValued Senior Member

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WTF? I have not "come around in this post only". You are simply refusing to listen.
What I have disputed was your generalised claim in post #86: "So a terminal speed cannot be achieved unless the horizontal component becomes zero first, as long as horizontal speed is present the bullet will not achieve terminal speed."
The first part of the sentence is the general principle you were asserting, the latter part being that principle applied to the bullet.
In post #101 you even continue to discuss the general principle you asserted.
Everything else I have said since is in relation to this - and it can be (however unlikely) applied to the issue of the bullet, should the bullet have aerodynamic properties.

If so then the bullet could very well demonstrate behaviour such that the terminal velocity need not be when the horizontal velocity is zero, even if only marginally above zero. I have set out above how this could be achieved. Are you disputing the physics? So do you accept that if the bullet had such aerodynamic properties that you would be wrong in your claim that the bullet would not achieve terminal velocity until its horizontal velocity was zero?
Again, I have not "just come around". Yes, I am disputing your general principle as you set out in post #86, and the fact that the argument against the principle as I have laid out can, however unlikely, be applied to any object that has an aerodynamic profile.
If you wish to be so dishonest as to close your eyes to what others have to say then do so, and be treated accordingly.

21. ### BaldeeeValued Senior Member

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Drag is a vector force.
Drag is proportional to velocity^2.
Assuming the same coefficient of drag and x-sectional area then the drag forces of the bullet travelling 824 m/s on a 200/800 trajectory will be identical to the sum of the drag of a bullet traveling 200 m/s horizontally and a bullet traveling 800 m/s vertically.

Why do you think this wouldn't be true?
This notion has underpinned applied physics for a good many years.

22. ### James RJust this guy, you know?Staff Member

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The magnitude of the drag force depends on the square of the speed of the object. The object's velocity has, in general, three vector components. The drag force acts in the opposite direction to the velocity at all times.

It is not true, in general, that the vertical component of the drag force is proportional to the vertical component of the velocity, because of the dependence of the drag force on the speed rather than the velocity.

This is a somewhat subtle point. Perhaps an example is in order.

Suppose the velocity of an object is $v=i-j$, where $i,j]$ are unit vectors in the horizontal and vertical directions at a given instant of time. The speed in this case is $\sqrt{2}$ units, and the drag force is proportional to the speed. The drag force vector is therefore proportional to $\sqrt{2}(-i+j)$.

The vertical component of the drag force at this instant is proportional to $\sqrt{2}j$, but the vertical velocity at this instant is $-j$. Compare this situation to the one in which the object is falling vertically, with velocity $v=-j$ (i.e. no horizontal component). In that case, the vertical drag force will be $j$ (without the $\sqrt{2}$ factor). So, you can see that the horizontal velocity component affects the vertical drag force.

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23. ### RajeshTrivediValued Senior Member

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James R has already explained this. Pl refer to my #127 where I had clearly stated that let V2 term not create any confusion. The complexity is that even the drag coefficient is speed dependent, and of course cross section for a bullet is different in both the directions.