The effect of gravity on a speeding bullet

Agree

Fire a bullet at same moment as dropping one

Both hit ground same time

MythBusters did this if you can find the clip

 

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No, origin, if the horizontal velocity component that too as high as 200 meter/second is present, then it cannot be ignored. This is the crux (along with sufficiency of 40000 ft) which I wanted to convey to you...and later on to kittamaru.

Let me highlight. The terminal velocity will be acquired only when Gravitational Force (downwards) is equal to the air drag force (upward). Force is a vector quantity, gravitational force is downwards, but due to presence of horizontal velocity component the the resultant drag force will not be in exact opposite direction to gravity. 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.

 

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Agree

Fire a bullet at same moment as dropping one

Both hit ground same time

MythBusters did this if you can find the clip

I found the clip

https://www.youtube.com › watch

 

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Cancelled

 

https://www.google.com.au/amp/s/www...ters-bringing-on-the-physics-bullet-drop/amp/

Perhaps the above link will help

MYTHBUSTERS: BRINGING ON THE PHYSICS BULLET DROP

 

Terminal velocity is generally understood to be one of either the vertical or horizontal elements of the speed. Objects can happily reach a vertical terminal velocity while maintaining a non-zero horizontal velocity. It depends upon the aerodynamics of the object in question, especially aerofoils. So it is not true that the horizontal component needs to be zero.
However, it is possible (I am not sure) that horizontal terminal velocity would be achieved at the same time as vertical terminal velocity, at least in a consistent fluid.

While the horizontal velocity might affect the rate of change in velocity (again taking into consideration aerodynamics), this merely informs the function for the rate of change of vertical velocity within the equation for establishing what that vertical terminal velocity will be, and when it is reached. If you have that function then everything else regarding the horizontal seemingly can be ignored, assuming that the horizontal is not a limiting factor (I.e. A brick wall a certain distance away that might curtail all movement, just as the ground is for the vertical).

 

Very entertaining.
However, extrapolating from a 45 round to all bullets seems problematic.
"39.6 millisecond difference..... no drop in fired bullet for 20 feet", then, How long did it take the fired bullet to move 20 feet horizontally?
Could that be the 39.6 millisecond difference?
if so,
then
a smaller projectile traveling at a much faster speed would seem to most likely have a longer hang time before dropping.

(all is a guess)

The muzzle velocity of the 45 is about 830 feet per second.
Which equals traveling about 32 feet in the mentioned 39.6 milliseconds

The above named(300 win mag) bullet would leave the barrel at about 3000 ft/second and have no appreciable drop for a tad over 100 yards. It would cover that distance in about 100 milliseconds.

(I ain't completely confused
but i am working on it)

............................
It would seem:
"Close, no cigar"

 

OP also gives no indication of velocity... thus, for the sake of simplicity, it is best to ignore horizontal velocity. Unless, of course, you wish to fill in specifics (speed of aircraft, exact altitude, barometric pressure, humidity, weather patterns, wind speeds, type of gun used, caliber and grain load of bullet, etc).

If you want to obfuscate this, trust me - it can be obfuscated into oblivion.

They are outside the scope of the question asked... no specifics were asked, and no details were provided. Thus, you would be attempting to solve math with ZERO constants and an equation full of variables for which we have precious few numbers to insert. What would be the point, without putting in your own data?

He stated terminal velocity would be achieved. He explained why (gravity, air resistance, et al). You are claiming this to not be true... so, prove it to not be true. If anything, as the one making the counter-claim that is opposing common sense logic, the onus is on you to provide some evidence that his claim is false.

Simply put: If you do this with a 120mm howitzer round or a .22 caliber rimfire cartridge, you will get two VERY different results.

 

But your drag vector can be decomposed into a horizontal component and a vertical component. These two are orthogonal and independent. The horizontal component has no effect on vertical motion. The term "terminal velocity" relates to the velocity of falling. So any horizontal component of velocity does not contribute to terminal velocity. Surely?

 

Everything near the surface of the earth will accelerate at 9.8 m/s^2. So the instant a bullet leaves the muzzle it begins accelerating towards the ground. Whether we are talking about a huge bullet or a tiny bullet it will feel an acceleration of 9.8 m/s^2. Even a feather when dropped accelerates at 9.8 m/s^2, however the acceleration is almost immediately slowed by air resistance so the feather floats slowly downward (slow terminal velocity).

A smaller bullet may be more influenced by air movement so it could be have a longer or shorter hang time, depending on the air movement.

 

You are tinkering with the definition.
When the terminal velocity is acquired the net acceleration on the object is zero. Now if the horizontal component of speed is present, it will have certain drag acceleration which will change the resultant speed of the object.

There is another issue here, offhand I am not sure if I am right in my conclusion on that, the drag force is in opposite direction of the motion of bullet, say at any instant the speed is (100,500) that is downward component is 500 meter per seconds and horizontal component is 100 meter per seconds, the drag force shall be at an angle arctan(100/500) from vertical direction away from earth. Now this drag force can be split into two orthogal components, one in vertical direction and another in horizontal direction. My conclusion is (I may be wrong here, some one may correct) that due to extreme non linearity of drag force, the vertical drag force with (100,500) will not be same as drag force with (0,500).

 

Could be in the realm of acceptable discrepancy error??

Also from me a guess

 

????!!!!

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I think I will leave this matter to others to take up, if they want to.

 

I'm with you

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Real good reason to repeat the process again and again and ..............
Then, go with the average?

 

Agree yes

Or until get it perfect?????

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Ain't nobody's perfect.
Just striving for a bit more accuracy is enough.

 

Way to go

 

Before it almost went of control, this is what I said, wish you had interjected earlier.