Speed of a falling bomb in ww2?

Answer 1

I found an Aviation History site with a paper written in 1924 that mentions the speed (the terminal velocity) of bombs, as this was important to compute and test to be able to aim bombs when dropping them from moving aircraft. The terminal velocity is the maximum speed at which an object will fall before air drag prevents it from dropping any faster. This of course varies depending on the weight and aerodynamic shape (the air drag coefficient) of the bomb. The altitude of the aircraft also determines whether the bomb has time to reach its terminal velocity before hitting the ground.

The terminal velocity for dropped conventional bombs varies from about 600 to 1000 feet per second (409 mph to 682 mph) (658 kph to 1097 kph).

Answer 2

Depends on drop method, air resistance, and drop altitude. A bomb from a dive bomber would be much faster since it is "launched" at the diving aircraft's speed. Dropped from a conventional bomber, at some speed, air resistance will balance acceleration and the bomb will reach terminal velocity. This will take much longer to happen with a bomb from a dive bomber. In either case, the bomb must be released high enough to allow time to reach terminal velocity.
Sorry I can't give you actual numbers.

Answer 3

v= v(initial) + at works in a vacuum in which WWII bomber did not fly. v=obviously velocity. a= acceleration

D= 1/2 CpAv^2 Where C=Drag Coefficient, p= rho=air density, A=crossectional area v= obviously velocity.

However at some some value of v Drag will equal mg this point is called terminal velocity. m=mass, g=gravity

so, 1/2CpAv^2 = mg

v= (2mg/CpA)^(-1/2)

This is only a rough formula to calculate the terminal velocity of the bomb because aerodynamics get very complicated, but this should get you somewhat close. All you need is the mass of the object, gravity which is 9.8 m/s^2, the drag coefficient, air density at the point that you want to know presumably fairly close to sea level, and the cross sectional area.

Answer 4

At which altitude? like 25k ft, or at the ground 0 ft? or would you like a formula for any point?
The bomb would fall from 0 ft/s and accelerate with gravity at 32. ft./s/s , minus a very slight coeffecient of friction from air resistance over the material covering the bomb and area of the bomb.
so, our formula would be something like d=(vo)t + (1/2at2 - F)
d= distance vo=the velocity the plane was rising or falling t=the time it took to hit the ground a=gravity F=friction (f=un)
something like 1285 ft/s at the ground if it was dropped from 25k ft. or like 2.5 mach. thats fast... i calculated like 1700 mph assuming the 25000 ft, and they probably didnt drop from that high....

Answer 5

Any object accelerates down with 9.8 m/s2. It means every second during fall the speed increases with 9.8 m/s

Answer 6

~as long as it was just dropped, it would fall at the same rate as a tiny pebble. The laws of physics and gravity say that any form of falling matter falls at the rate of 9.8 meters per second.

Answer 7

Well yes, its acceleration would be 9.8m/s^2 but keep in mind it may hit a terminal velocity before hitting the ground. However I'm sorry to say I really don't know what that would be.

Answer 8

The same speed at which money flies out of my pocket.

Answer 9

depends on the height it was dropped from and current height at an acceleration of 9.8 metres per second per second

Answer 10

423 mph or so