Question:
During WWI, the Germans had a gun called Big Bertah that was used to shell Paris. THe shell ahd an initial...?
drdagher89
2006-09-21 17:47:58 UTC
speed of 2.6 km/s at an initial inclination of 75.9 deg to the horizontal.
a) How far away did the shell hit? Answer in units of km.

b) How long was it in the air? Answer in units of s.
Four answers:
anonymous
2006-09-21 17:52:56 UTC
Big Bertha

Updated - Saturday, 2 August, 2003



Although the name was commonly applied to a whole variety of large-calibre German artillery guns the "Big Bertha" ('Dicke Berta') actually referred to a single siege gun, at that time the world's largest and most powerful.



Produced by the German firm of Krupp the Big Bertha was a 42cm howitzer, model L/14 designed in the aftermath of the Russo-Japanese War of 1904 on behalf of the German Army. It was initially used as a means of (successfully) demolishing the fortress towns of Liege and Namur in August 1914, the war's first month (and subsequently as Antwerp). It was thereafter used to similarly reduce other enemy strong-points as the need arose.



The somewhat unflattering name itself arose from association with the wife of Gustav Krupp, owner of the Krupp factory. Her name was Bertha Krupp von Bohlen und Halbach.



Only four Big Bertha howitzers were produced, the first two rolling off the production line a mere matter of days after the onset of hostilities, on 9 August 1914. Once constructed these huge guns, whose shells weighed 820kg each, were shipped in their constituent parts by tractor to their destination point where they were once again reassembled by a huge crew of as many as 1,000 men.



With a range of 15km their 420mm shells proved devastating and all four were used during the German assault upon Verdun from February 1916.



Once the Verdun offensive was called off in failure (leading to the replacement of German Chief of Staff Erich von Falkenhayn who had initiated the battle) the Big Bertha guns were decommissioned, since Allied artillery developments had resulted in guns with a longer range.
anonymous
2006-09-22 01:49:12 UTC
I guess you want some calculations. OK. Take in consideration that air resistance is insignificant OK? So we have horizontal velocity = 0.633 km/s (this value was calculated using the Pythagorean theorem.) Where sin 75.9 = horizontal velocity / 2.6 km/s. we solve for horizontal velocity and we get 0.633 km/s.



Now we also need to find vertical velocity which again using the Pythagorean theorem is equal to 2.52 km/s. so far so good? Now we have to find the maximum height reached by the shell for that we use this equation:



(Vertical velocity) ^2 / 2g = maximum height. Where g = 9.8 m/s

^2 (In this equation you have to convert vertical velocity to m /s) vertical velocity = 2520 m/s



With numerical values we have



2520^2 / 2*9.8m/s^2 = h therefore h =324000 m



Now compute the time it would take the shell to hit the ground at that altitude



We find out that the time taken is 257.14 s but remember that the total time is



Double because it took time to get to the maximum height and it took the same time



To get down therefore we have that the total time the shell was in the air was



514.28 s.



Now answering your questions



a) 325.54 km



b) 514.28 s
Dennis K
2006-09-22 00:58:27 UTC
From what city were they firing the gun, there are know air currents in Europe to deal with. You must also take into account air temperature and resistance which decrease with altitude and barometric pressure. What time of year was it during the firing, I will look up the weather patterns and get you a close answer.
mekaban
2006-09-22 08:38:07 UTC
Let V= initial velocity at 45 deg, Vv= vertical velocity, Vh=horizontal velocity, t1=upward time, t2= downward time, h1= vertical height, and R=horizontal distance. Then,



Vv=2.6*Sine 75.9

Vv=2.52km/s

Vh=2.6*cosine 75.9

Vh=0.633 km/s

h1=Vv^2/(2*g)

h1=(2520m)^2/(2*9.81)

h1=323670 meters=323.67km

t1=Vv/g

t1=2520/9.81

t1=257 seconds

t2=t1=257 seconds

R=Vh*(t1+t2)

R=0.633*(257+257)

R=325.4 km

Therefore,

a)It hit 325.4 km away

b)It was in air for approximately 514 seconds.


This content was originally posted on Y! Answers, a Q&A website that shut down in 2021.
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