How Far Did Big Bertha's Shell Hit During WWI?

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Homework Help Overview

The discussion revolves around a physics problem related to projectile motion, specifically analyzing the trajectory of a shell fired from the Big Bertha gun during World War I. The original poster presents initial conditions, including speed and angle, and seeks assistance in setting up the calculations to determine the distance traveled and time of flight.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the decomposition of the initial velocity into horizontal and vertical components using trigonometric functions. There are questions regarding the correctness of calculations and the setup of equations for motion.

Discussion Status

Some participants have provided guidance on calculating the components of velocity and suggested methods for determining the time of flight and distance. There is an ongoing exploration of the correct approach, with participants verifying calculations and discussing assumptions about the motion.

Contextual Notes

Participants mention the use of a calculator program and the importance of ensuring the calculator is set to degrees for accurate trigonometric calculations. There is also a focus on the need to clarify the setup of equations for both horizontal and vertical motion.

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During World War I, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 1.11 km/s at an initial inclination of 35.8 degrees to the horizontal.
a.How far away did the shell hit? Answer in km.
b. How long was it in the air?

I have a RECTPOLAR program on my calculator that we are allowed to use. I know I need the resultant and the angle for that to give me the x magnitude and the y magnitude. How would I go about setting this up?
 
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Vx = V.cos(35.8) horizontal component

Vy = V.sin(35.8) vertical component
 
i don't think that makes sense...
 
cos(35.8) gives me -.32 and sin(35.8) gives me -.95

i don't think that works
 
Make sure you have your calc. set to degrees ( as opposed to radians).
Cos(35.8) = 0.81106.
 
ohh ok now i got that...

so vx = .811 and vy = .585

so should i go and do the vf, vo, a, d, t set up?
like
vx .811
dx?
t

vfY
voY 0
aY -9.8
dY
t

and if so, how would i find a 3rd component to the Y part?
 
Not quite. Look at post #2 again.

Vx = 1.11*0.811 Km/s
Vy = 1.11*0.585 km/s

Using Vy you can work out how long the shell is in the air ( think of a stone thrown vertically with Vy ). The time taken to get to where Vy=0 is half the travel time. When you know the travel time, multiply by Vx to get the distance.
 
Last edited:

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