# how long did it take the bombs to hit the ground from 25 000 ft



## mikey (Dec 10, 2013)

I was wondering what the avg bomb drop height of the European theater and how long it took to hit the ground thanks in all the movies it shows them hitting right after release


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## pattle (Dec 10, 2013)

I haven't a clue but they don't just fall straight down the motion of the aircraft throws them forward.


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## Greyman (Dec 10, 2013)

Depends on the bomb, speed of aircraft, wind, etc. but roughly:

5,000 - 17.9 sec
10,000 - 25.5 sec
15,000 - 31.5 sec
20,000 - 36.6 sec

(RAF GP 250 lb)


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## mikewint (Dec 10, 2013)

Depends on multiple factors. In the simplest sense Distance traveled under uniform acceleration = (.5)(acceleration due to gravity)(time)^2 power. Using your 25,000ft height and the average gravitational acceleration of 32ft/s^2 (varies with height and latitude). *So 39.5s *BUT all falling objects IN AIR reach a terminal velocity wherein the downward force tending to accelerate = the upward force of air resistance. Air resistance is a very tricky and impossible to calculate exactly because air density varies with height and temperature of the air, plus we have to consider the cross-sectional area of the bomb and its surface texture and fins. 
In the simple example above air resistance was ignored which also would mean that the bomb would continue to accelerate over the entire distance hitting the ground at 862MPH!! Since the speed of sound is roughly 750MPH the bomb hits at MACH 1.2, obviously not true since you can hear bombs as they fall.
So the best estimate I can give you is a terminal velocity of between 550MPH and 650MPH or between 46.5s and 39.3s
ALSO the horizontal velocity of the plane has NOTHING to do with the time of vertical fall. The two motions are INDEPENDENT of each other. As in crossing a river by heading directly across. The river's current determines where you reach the opposite bank

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## parsifal (Dec 10, 2013)

In a vacuum, the accelaration of any object is 9.8m/s. however in atmospheric conditions, the object will reach terminal velocity, and this will vary according to the density of the atmosphere. At high altitudes, where the atmosphere is thin, there will be a higher terminal velocity, at lower altitudes, the terminal velocity is lower. if the atmosphere was so thick it was a liquid or even solid, the terminal velocity would be very low, or close to zero.

Terminal velocity is where the downward force of gravity equals the air resistance/force of drag and buoyancy, velocity becomes asymptotic to a constant value with net acceleration .

Typically, a skydiver has a terminal velocity of 195 kmh, but this can vary a lot. The world record rate of descent is about 800KmH. A standard rifle bullet has a vertical rate of descent of about 300 KmH or 90m/s

one would expect an iron bomb, with a similar shape to a bullet, would have a similar terminal velocity in the vertical. One would expect the distance travelled by the ordinance to be more than the vertical distance to the ground, since it retains some horizontal movement as well. But its difficult to determine how much of that forward movement is retained, and for how long. If we assume that for a drop at 20000ft the bomb actually travels 25000 ft, or 8000m, at a terminal velocity of 320 kmh, it should take about 90-95 secs for the bomb to make impact.

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## Shortround6 (Dec 10, 2013)

Bomb and bullet may have a similar shape, what they don't have is similar sectional density (weight per unit of frontal area), which is why artillery shells travel soooo much further than rifle bullets even with near the same shape and fired at near the same speeds.


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## parsifal (Dec 10, 2013)

I get that, but the terminal velocity is still the issue. What is the terminal velocity of a standard 500 lb GP bomb for example. Once we have that, we will be able to get stuck into this problem.


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## fubar57 (Dec 10, 2013)

From(shudder) Wikipedia Bombsight - Wikipedia, the free encyclopedia

The M65 will be dropped from a Boeing B-17 flying at 200 mph at an altitude of 20,000 feet in a 25 mph wind. Given these conditions, the M65 would travel approximately 6,500 feet forward before impact,[8] for a trail of about 1000 feet from the vacuum range,[9] and impact with a velocity of 1150 fps at an angle of about 77 degrees from horizontal.[10] A 25 mph wind would be expected to move the bomb about 300 feet during that time.[11] The time to fall is about 37 seconds.

Geo


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## mikewint (Dec 11, 2013)

Parsifal, The actual path distance traveled by the bomb is immaterial to this question as the motion is in two planes and the two motions are independant of each other. Thus the vertical velocity is exactly the same whether the object is dropped from a moving aircraft or a stationary cliff. A bullet frired horizontally and a bullet dropped from the barrel at the same instant reach the ground at the same time. Your time of 90s is vastly too long. 
Air resistance varies as the cube of the cross-section thus a bullet and a bomb will not have a similiar terminal velocity.
Arty shells travel further than bullets because they are fired above 45 degrees into thinner atmosphere thus reducing air resistance to the horizontal velocity while the added upward vertical velocity increases time of flight (range). Arty shells fired north or south are also affected by corollis forces producing curved paths


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## GregP (Dec 11, 2013)

In a vacuum, the bombs would fall straight down from the bomb bay and stay right under the bomb bay as the aircraft moves forward. It would accelerate at 9.8 m/s^2 but the horizontal velocity would remain unchanged. Let’s say the B-17 is bombing from 25,000 feet at 180 mph and somehow is doing it in a vacuum.

180 mph is 80.46 m/s and 25,000 feet is 7,620 m. The bomb continues to move in the x (horizontal) direction at 80.46 m/s and accelerates from an initial value of zero in the y (vertical) direction to a final value of 386.46 m/s downward and impacts the ground in 39.435 seconds. 346.46 m/s is 864.5 mph downward and 180 mph forward. The combined speed of the bomb would be 883 mph in a vacuum.

In the real world, the plane cannot fly in a vacuum; it flies in air. The real secret to the Norden bombsight, or any other bombsight, is that it computes the effects of air resistance and shows the point of impact of the bomb continuously in the sight. When the target shows up in the crosshairs, press the release. The bomb’s terminal velocity in real air would vary with altitude, but gets slower as the bomb gets closer to the ground because the air gets denser as you approach the ground.

In the figure below, the solid line is how the bomb would fall in a vacuum. It will stay right under the bomb bay as it descends and gains vertically velocity. In real life, it follows the approximate path of the dotted line. The amount it falls shorter than vacuum is determined by the altitude of release, the cross-sectional area of the bomb, the drag of the bomb, and any winds that the bomb falls through on the way down.







I doubt of the real bomb will get to 860+ mph and would guestimate the real speed to be in the 600 – 650 mph range depending on bomb type. I’d not be surprised to be off either high or low. I have books on artillery fire that explain a lot, but they are all several hundred pages of math. 

Since the vertical velocity decreases by about 1/3, I’d expect the fall time to be in the 59 second range from 25,000 feet if I try a terminal velocity of, say, 600 mph. I'd therefore expect anywhere from 50 seconds to about 1 minute 10 seconds as a first-order estimate.


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## CobberKane (Dec 11, 2013)

> Parsifal, The actual path distance traveled by the bomb is immaterial to this question as the motion is in two planes and the two motions are independant of each other. Thus the vertical velocity is exactly the same whether the object is dropped from a moving aircraft or a stationary cliff. A bullet frired horizontally and a bullet dropped from the barrel at the same instant reach the ground at the same time.



A lot of people really have trouble getting their head around this. I've spoken to experienced shooters who point blank refuse to accept the above example. Gotta admire their self belief, backing their opinions against Newton in this area!

Regarding the terminal velocity of bombs - weren't the tallboy and grand slam supersonic? if so, there must be quite a range according to the purposes of various bombs, from incendiary to armour piercing


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## GregP (Dec 11, 2013)

Not sure if the tallboys were supersonic or not, could be so and have heard that in the past with no references to support it, but there certainly WAS a wide range of drag and shapes in bombs, making a single bombsight a complex item. Looking at the shape of a tallboy, it might easily be possible that it was supersonic given the weight and a reasonably high drop altitude. Whether they were supersonic or not, their mass would certainly impart tremendous momentum, p (p = mv), making their penetrating power impressive. The mass was 10.1 times (an order of magnitude higher) that of a 2,000 pound GP bomb which was, in itself, an impressive instrument of destruction.

Many ships were sunk with 500-pound bombs.


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## stona (Dec 11, 2013)

The maths is here for anyone who is interested. Greyman's 25.5 sec from 10,000 feet is pretty much the same as the exemplar used in this exercise.

http://msemac.redwoods.edu/~darnold/math55/deproj/sp09/wilsonpepper/cpandcwfinal.pdf

Extensive tests were done using dummies of the nuclear weapons to be dropped on Japan. Not exactly GP bombs but the variations are less than you might expect. The results are tabulated here.






Cheers

Steve


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## Greyman (Dec 11, 2013)

Sorry, I should have sourced my figures, it might have sped things along here. They're from '_Air Publication 1243 Armament Training for the RAF - Part II - Bombs, Pyrotechnics, Bombsights, etc._' and can probably be regarded as reasonable accurate.


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## mhuxt (Dec 11, 2013)

There's an altitude vs. time-of-fall graph in this article:

http://legendsintheirowntime.com/Content/1943/Fl_4310_bombsight.pdf

Numbers seem to tie in to Greyman's OK.

Have various other bits and pieces, will try to upload tomorrow, beddy-byes now, kid wants to go running in the morning...

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## mikewint (Dec 11, 2013)

Greg, you need to consider only the bomb's vertical velocity which is zero at time of release. It is also unnecessary to convert to metric measures. Average g over the US is 32ft/s^2 (as I stated earlier it varies with height and latitude). At bomb release Vvert = 0. Once released the bomb will accelerate negatively at approximately 32ft/s for every second of fall opposed by air resistance (impossible to know exactly). The bomb will accelerate (at a decreasing rate until a = 0) to it's terminal velocity (also not constant over the entire fall as air resistance is constantly changing). Once at terminal velocity it will fall at a constant rate until impact.
Without more definitive date we can only make assumptions. Assume the bomb accelerates for 1/2 the distance or 12,500ft then Vvert goes from 0 to 550mph over 12,500ft. The time required to cover this distance is the same as if the bomb had covered this distance at the average velocity of 275mph. Since D = vt; t = 30.99s The bomb then covers the remaining 12,500ft at the terminal velocity of 550mph requiring t = 15.496s. This gives a *total time of 46.5s* Using the same logic for a more streamlined bomb and a terminal velocity of 650mph we get t = 26.224s and t = 13.112s or a *total time of 39.3s*
Thus the extremes are 46.5s and 39.3 or most likely the average of the two about *42.9s*
As a note, the worlds first programable computer was ENIAC (Electronic Numerical Integrator And Computer) its main function was to calculate arty firing tables. It simply took too long for humans to do all the massive calculations required to predict the landing point of a shell


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## stona (Dec 11, 2013)

mikewint said:


> As a note, the worlds first programable computer was ENIAC (Electronic Numerical Integrator And Computer)



And G.P.O. engineer Tommy Flowers gets overlooked again 

The Mk 1 Colossus ran in November 1943 for the first time. I don't want to start a row about what the term 'programmable' means 

Cheers

Steve


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## mhuxt (Dec 11, 2013)

The wiki entry for the bombsight mentions a "Terminal Ballistic Data" file, which I've tried to upload in this post, but the site boots me out, perhaps due to size of the file. It looks like the very highest speeds achieved by US bombs from 35,000 feet are just in excess of 1,200 feet/sec, doesn't seem to vary greatly with weight of bombs.

I have another pdf file with weapons data, which also has terminal velocity curves, maxing out at about 1,050 ft/sec for 500 lbers and 1,075 ft/sec for 1,000 lbers, when dropped from 32k feet.

There's a bombardier's information file at this rather interesting site:


View Items

It has some Navy ATF data, and a small copy of a bomb table referred to when setting data into the Norden, on which I've highlighted relevant time of fall and altitude entries.

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## mhuxt (Dec 11, 2013)

For what it's worth, I had a look at my other pdf file (43-odd megs, watermarked to USSBS.com) for Grand Slam and Tallboy data, apparently the GS had a striking velocity from 30k feet of 1,375 feet/sec, which apparently converts to 937.5 mph, so yes, supersonic.


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## GregP (Dec 11, 2013)

Nice paper mhuxt!

The problem is determining the cross-sectional area and all the constants, but it is a pretty decent read.

Thanks!

I was using a text on artillery fire and assuming a zero angle of elevation and the height as distance below the start point where the target is located. It gives the same results.

Hi Mike!

I don't think so ... if you consider only the vertical velocity, you will fail to take into account for the slowing of the horizontal velocity by air resistance and will fall short of you aim point. The TIME OF FALL would be correct if you know the terminal velocity and the equation of it, but the HORIZONTAL TRAVEL DISTANCE would be wrong by a a bit and you'd impact short of target.


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## mikewint (Dec 11, 2013)

mhuxt, from the data table you supplied the Tall Boy dropped from 25,000ft strikes at 1260ft/s BUT at an angle of 16 degrees. While the table does not specify this it must be 16 degrees from vertical. Thus we have a right triangle where 1260 is the hypotenuse. This resolves itself into two motions a vertical (y) and a horizontal (x) velocity. Using trig functions: sin(16) = x / 1260 or x = 347ft/s (236 mph) and cos(16) = 1211ft/s (826 mph) which is still supersonic. However,
In John Ellis book "One Day in a very Long War",: Tallboy was designed to be dropped from an optimal altitude of 18,000 ft at a forward speed of 170 mph. *Impacting at 750 mph *. Not supersonic


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## mhuxt (Dec 11, 2013)

Gents,

Couple more pages from the same file as has the Tallboy/Grand Slam info.


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## mikewint (Dec 11, 2013)

GregP said:


> Nice paper mhuxt!
> 
> The problem is determining the cross-sectional area and all the constants, but it is a pretty decent read.
> 
> ...



GregP, I am not sure to what you are refering too. A bullet/shell begins to accererate at -32ft/s/s the instant it leaves the barrel dropping 16ft in one second with a vertical velocity of 32ft/s. How far down range the impact occurs depends upon the horizontal velocity. The horizontal velocity does not have a "terminal velocity" as such unless you consider 0 to be a velocity. Gravity does not act horizontally thus the initial horizontal velocity constantly decreases due to air resistance (friction). Given sufficient time this could become zero with the bullet/shell simply falling straight down. Anyone who shoots at distant targets must know their bullet drop. 
For the M16 a .223 round drops 2in in the first 100yds; 8.7in in the next 100yds; and 41.9in at 400yds. The increasing drop is due to the ever decreasing horizontal velocity and the ever increasing vertical velocity. An M16 round takes about a second to travel 650yds at which point it has dropped 12ft 
The average fall of the earth's surface is 1ft per mile thus an object moving fast enough can cover that mile in the time it takes to fall 1ft (.25s) or 14,400mph. Thus at the end of 1s of fall the object is still at the same height and so on as the Earth's curvature is followed, i.e. you are in orbit

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## mhuxt (Dec 11, 2013)

Sorry, should have included the 500 lb SAP bomb data.


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## parsifal (Dec 11, 2013)

man that is some exceelent legwork. well done


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## GregP (Dec 11, 2013)

Hi Mike,

In a vacuum, the bomb drops at constant horizontal speed and accelerates according to V = at where a = 32 ft/sec^2 or 9.8 m/s^2. In the air, the horizontal speed decays with time as the bomb drops and it accelerates vertically much more slowly than in vacuum until it impacts or reaches terminal velocity. So the horizontal speed is variable and continuously slowing while the vertical velocity is increasing until either impact or terminal velocity is reached.

Whay I am saying simply is that if you consider the vertical component alone, then you will misss horizontally by some amount since the horizontal velocity is NOT constant. Probably not by too much, but easily enough to negate a LOT of damage from 15,000+ feet.

This all assumes ONE bomb is dropped.


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## wuzak (Dec 12, 2013)

mikewint said:


> mhuxt, from the data table you supplied the Tall Boy dropped from 25,000ft strikes at 1260ft/s BUT at an angle of 16 degrees. While the table does not specify this it must be 16 degrees from vertical. Thus we have a right triangle where 1260 is the hypotenuse. This resolves itself into two motions a vertical (y) and a horizontal (x) velocity. Using trig functions: sin(16) = x / 1260 or x = 347ft/s (236 mph) and cos(16) = 1211ft/s (826 mph) which is still supersonic. However,
> In John Ellis book "One Day in a very Long War",: Tallboy was designed to be dropped from an optimal altitude of 18,000 ft at a forward speed of 170 mph. *Impacting at 750 mph *. Not supersonic



I don't think that the Tallboy was designed to be dropped from 18,000ft but that is how high the Lancaster could carry it. It was designed to be dropped from much higher altitudes, 30-40,000ft, IIRC.

As far as the speed goes, I would think the speed of the Talboy along its axis would be more important for its effectiveness than either the horizontal or vertical components. You could argue that those components are vital for aiming the bomb, butthe release mechanism was not the most reliable system (supported by a chain which is released to drop the bomb).


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## GregP (Dec 12, 2013)

Hi Wuzak,

I'm curious. Since the Lancaster hauled the largest payload in the ETO, what possible difference could it have made what altitude the Tallboy was made to drop from? The ONLY imporant thing here is what altitude it WAS dropped from. If 18,000 feet was it, then that was the best that could be done unless the ETO suddenly acquired B-29's, which it never did.

Not arguing here ... just saying, if there is no platform to carry the thing to 30,000 feet, then what possible use is it to say the design altitude was 30,000 feet? It's like saying "My cheating wife would have remained true to me if only she hadn't cheated" or "maybe I'll have a beer ..."

The claim is pointless, even if accurate.

I WILL have a beer now. Cheers!


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## wuzak (Dec 12, 2013)

Barnes Wallis had also proposed the 6 engined "Victory bomber" to carry such earthquake bombs. Whether or not that could have carried the bombs to teh required altitude is another matter.

Wallis reworked his calculations to take into account the lower altitude at which the bombs could be carried and check to see if the bombs would still be effective.


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## stona (Dec 12, 2013)

The Bielefeld viaduct was attacked from only 12,000 feet. One Grand Slam dropped by Squadron Leader Calder did what over 3000 tons of conventional bombs had failed to do. The time recorded for the fall of the bomb was 35 seconds though I have no idea how accurate that might be.
Cheers
Steve


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## Shortround6 (Dec 12, 2013)

GregP said:


> Hi Wuzak,
> 
> I'm curious. Since the Lancaster hauled the largest payload in the ETO, what possible difference could it have made what altitude the Tallboy was made to drop from? The ONLY imporant thing here is what altitude it WAS dropped from. If 18,000 feet was it, then that was the best that could be done unless the ETO suddenly acquired B-29's, which it never did.
> 
> ...




Rather misses an important point which is relevant to this discussion. _IF_ you need a bomb with the weight to frontal area ratio _and_ the shape of a Tallboy/Grand Slam to go supersonic then the chances of a "normal" WW II shaped bomb reaching such a speed is about zero. 
The Fact that the Tallboy/Grand Slam didn't achieve supersonic speeds in service due to limitations of the carrying aircraft _does not_ mean that "normal" bombs could. The Tallboy/Grand Slam _design_ puts an upper speed limit on what was achievable in WW II.


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## pbehn (Dec 12, 2013)

Would any free falling bomb pass through the sound barrier if dropped from within the atmosphere? Or would the shock waves make a natural limit?


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## drgondog (Dec 12, 2013)

mikewint said:


> mhuxt, from the data table you supplied the Tall Boy dropped from 25,000ft strikes at 1260ft/s BUT at an angle of 16 degrees. While the table does not specify this it must be 16 degrees from vertical. Thus we have a right triangle where 1260 is the hypotenuse. This resolves itself into two motions a vertical (y) and a horizontal (x) velocity. Using trig functions: sin(16) = x / 1260 or x = 347ft/s (236 mph) and cos(16) = 1211ft/s (826 mph) which is still supersonic. However,
> In John Ellis book "One Day in a very Long War",: Tallboy was designed to be dropped from an optimal altitude of 18,000 ft at a forward speed of 170 mph. *Impacting at 750 mph *. Not supersonic



Mike - the 'hypotenuse' is the actual velocity vector.. and if the target had an inclination of 16 degrees perfectly normal to the velocity vector the impact velocity would be the same as the velocity vector


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## mikewint (Dec 12, 2013)

GregP said:


> Hi Mike,
> Whay I am saying simply is that if you consider the vertical component alone, then you will misss horizontally by some amount since the horizontal velocity is NOT constant. Probably not by too much, but easily enough to negate a LOT of damage from 15,000+ feet.
> 
> This all assumes ONE bomb is dropped.



GregP, totally understood. If you read my posting I thought I had made that clear with the .223 bullet example. If you aim perfectly horizontally 6ft above the ground you cannot hit a target 400yds distant as the bullet drops 12ft in the time required to cover 400yds horizontally. So one aims ABOVE the target by the known bullet drop distance. Bomb sights, like the Norden take the aircraft factors into account for the bombardier. The theoretical CEP for the Norden was only 75ft though in combat it was more like 1200ft so pinpoint bombing was not possible thus the Navy turned to dive bombers and skip-bombing and the Air Force to the lead bomber technique.
My point was simply that you cannot consider ONLY horizontal or ONLY vertical as the bombs actual velocity is a composite (summation) of the two which is constanly changing as the bomb is accelerating (both + - ) over the entive time of fall. BUT it is the "time of fall" that determines range. The bomb/bullet can only move horizontally while it is falling. The aircraft can adjust horizontal speed + or - within limits. In todays supersonic bombers, release can be 20mi from a target


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## mikewint (Dec 12, 2013)

drgondog said:


> Mike - the 'hypotenuse' is the actual velocity vector.. and if the target had an inclination of 16 degrees perfectly normal to the velocity vector the impact velocity would be the same as the velocity vector



Agreed, I was trying to show that impact velocity is not the same as the vertical falling velocity nor the horizontal velocity but a summation of the two. Thus a bomb can impact supersonically (above 767mph) while the vertical vector and horizontal vectors are less than this

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## mikewint (Dec 12, 2013)

pbehn said:


> Would any free falling bomb pass through the sound barrier if dropped from within the atmosphere? Or would the shock waves make a natural limit?



You ask a "not-easy-to-answer" question since multiple factors are involved. We can start with the simple fact that the speed of sound is not a fixed number. It varies with the density, temperature, and composition of the conducting medium (air in this case). The density of air varies considerably with height as does temperature and composition. Thus a falling object can be both super and subsonic in its fall. Recently Felix Baumgartner did a freefall jump from over 24 miles and went supersonic (for a time) at around 30,000ft - 33,000ft (Speed of Sound = 761mph @ 15C @ sea level; 678mph @ 30,000ft @ -44C; 659mph @ 60,000ft @ -57C)
If you wish to fiddle around with the math: 
F(air resistance) = 0.5 A v^2 p(z) Cd Where A = max cross sectional area v = velocity Cd = drag coefficient p(z) air density @ z altitude
This is in turn equal to m g Where m = mass and g = 32ft/s^2 gravitational acceleration (varies with height but a few thousand feet over the earths radius is not significant.
Cd depends on shape and surface finish generally from 0.05 - 0.1 (tables exist for bomb type) This value is for the BOMB only, as the speed of sound is approached the shock wave formed adds exponentially to the drag.
So a simple answer is YES a simple one dimensional vertical drop can achieve supersonic velocities as Felix Baumgartner demonstrated (his Cd was a large 1 - 1.3).


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## GregP (Dec 12, 2013)

Hi Shortround,

Unsurprisingly we disagree, but maybe by not much. The Tallboy design allows supersonic impact but the available transport resources to carry it didn't allow for it since they could not get to the design height.

Maybe we're saying the same thing in different words.

The Lancaster could not reach the design height of the Tallboy, but I'd suppose that wasn't a huge comfort to those upon whom it was dropped. It was a BIG problem wherever it happened to impact, wouldn't you say?


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