aamc 5r PS #65 question

[russianguy13]

Have a question about AAMC 5R PS #65.
I was wondering if someone can explain to me why/how we can use the formula y=1/2gt^2 when in the equation it says the snowball was thrown from the ground level.

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[ljc]

Because ground level in this case isn't on flat ground. Ground level at the top is 20 meters above ground level at the bottom. Thus, when you throw it straight out from "ground level" at Point A, this is the same as throwing it off a 20 meter building. So y=20, G is always 10, and you solve for T, which ends up being 2. So 2*25=50m horizontal displacement, B is the answer.

Ignore the term "ground level" unless the terrain in question is ALL flat. If there is any discrepancy in height, use that as your metric instead of an arbitrary "ground level." You should almost always define the axes for yourself, so you won't get confused about where the author thinks you've defined them.

 

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