Exploding shell at the top of its trajectory

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After a shell explodes at the top of its trajectory, the fragments experience a downward acceleration equal to g, with no x-component due to the absence of air resistance. The discussion clarifies that the shell was fired straight upwards, which explains the lack of horizontal acceleration. The explosion does not alter the constant velocity in the x direction, as the fragments maintain their momentum. The confusion arose from the term "explodes," which led to assumptions about the direction of the fragments. Understanding the context of the shell's trajectory is crucial for grasping the physics involved.
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"After a shell explodes at the top of its trajectory, the center of gravity of the fragments has an acceleration, in the absence of air resistance, equal to g and downward."

Why isn't there an x-component?

I get why there would be a downward component with a = g, but shouldn't there at least be an increase in acceleration in the x dimension? Does it not have an acceleration in the x direction just because it's traveling at a constant velocity in the x direction (i.e., does it assume the explosion doesn't affect velocity in the x direction, albeit momentarily)?
 
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1/ it was fired straight upwards

2/ shaped charge setup to fire straight downwards

TBH it sounds like a really stupid question. If it "explodes" some portion of the fragments will expand outwards in all directions ...
 
d3mm said:
1/ it was fired straight upwards

2/ shaped charge setup to fire straight downwards

TBH it sounds like a really stupid question. If it "explodes" some portion of the fragments will expand outwards in all directions ...

Ah. My bad. I was assuming for some reason that it wasn't fired straight up. And yeah, I just wish it said split or something rather than "explodes". Anyway, I appreciate the help : D
 
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