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Old Jul 12th 2018, 09:43 AM   #1
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Homework question help

Hi everyone. I took calculus based physics years ago and I didn't do so well ( I went from 12 years in the army to trying to go into an engineering school) now many years after that, I am finishing another degree and my phys credits didn't transfer so now I am doing Algebra based physics. I am having trouble with the last part of this question maybe someone would be able to dumb it down for me? I do NOT need answers at least not with my numbers, I am more looking how to go about it and any relevant equations (I'm taking it online and I think online Physics should never be done).

The cannon on a battleship can fire a shell a maximum distance of 41.0 km.
I got the Vo and the Maximum height
Vo = 670.75
Hm = 10250

(c) The ocean is not flat, since the earth is curved. How many meters lower will its surface be 41.0 km from the ship along a horizontal line parallel to the surface at the ship?



Does your answer imply that error introduced by the assumption of a flat earth in projectile motion is significant here? (Select all that apply.)
The error is significant compared to the distance of travel.
The error could be significant compared to the size of a target.
The error is insignificant compared to the size of a target.
The error is insignificant compared to the distance of travel.


Thank to anyone who can help!
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Old Jul 12th 2018, 02:37 PM   #2
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Hi Edermardil ....this is the best diagram I could find that represents the situation , it's not very good but it will have to do, in our situation the angle theta will be very small ....



TP is the horizontal line projected from the battleship ...

We are told the max range of the shell is 41Km, this is the distance TP

distance OT is the radius of the Earth

distance OP is the radius of Earth plus the distance you are asked to find in question (c)...

so we have a right angle triangle ,simple geometry to find OP since we know the other two sides ...a search will get Earth's radius.
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Old Jul 12th 2018, 03:20 PM   #3
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Good answer Oz, and your method gets the same answer as my engineering solution.

0.1319 kilometres or 131.9 metres

I have come to the conclusion that the most common folmula in Engineering is (something squared) divided by twice something else.

In this case the something is the distance along the tangent and the something else is radius of the earth.
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