EPTInsane's answer is best. I never seen such an easy problem made so complex... you remind me of my math teacher back in high school LOL!
[H]Calculus - Page 2
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Rev0lution
United States1805 Posts
Insane's answer is best. I never seen such an easy problem made so complex... you remind me of my math teacher back in high school LOL! | ||
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infinity21
Canada6683 Posts
On October 18 2007 04:51 HonkHonkBeep wrote: This is not calculus That's exactly what I thought lol | ||
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fight_or_flight
United States3988 Posts
Where do you think those "magic" equations came from?  | ||
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Purind
Canada3562 Posts
On October 18 2007 10:49 Rev0lution wrote: that aint calculus. this is HS physics lol. Insane's answer is best. I never seen such an easy problem made so complex... you remind me of my math teacher back in high school LOL! Of course it's calculus. The physics equations that you use are derived from calculus, assuming constant acceleration. Equations of motions don't just pop out of nowhere, these ones came from calculus. | ||
Insane
United States4991 Posts
On October 18 2007 10:49 Rev0lution wrote: that aint calculus. this is HS physics lol. Insane's answer is best. I never seen such an easy problem made so complex... you remind me of my math teacher back in high school LOL! Obviously it's an easy question, I just derived (which is probably a poor word here since it was with integration, but whatever ) the equations, given the knowledge p'(t) = v(t), v'(t) = a(t) (and that a(t) is some constant value, here 9.8). You can apply the method to some other relationship, where for example, a(t) is some more complex function (say it's a(t) = t²). Then you can no longer use the basic kinematics equations which assume constant acceleration, and end up with some other more complex function. (in that case, a(t) = t², v(t) = (1/3)t³ + C, p(t) = (1/12)t^4 + Ct + C' , where C' is some constant which may or may not be different from C) It's a more general solution to a simple problem  | ||
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dancefayedance!~
396 Posts
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oneofthem
Cayman Islands24199 Posts
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Insane
United States4991 Posts
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GeneralCash
Croatia346 Posts
the formula is g = x" - fx' where f is related to friction. obviously, if f or h are small enough you simply integrate without -fx' from 0 to h and get the high school formula. but h is not small so the exponential term kicks in and the rock travels at the terminal velocity (which is constant) after falling long enough. without knowing the friction you can't do much. you might get the solution if you drop a rock form the edge of the building, then climb on a ladder of known hight and throw it again a few times (once should be enough i think). mesure the time but change the integral boundries when you use the ladders. | ||
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Insane
United States4991 Posts
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SpiritoftheTunA
United States20903 Posts
On October 18 2007 19:41 GeneralCash wrote: it's not as easy as it sounds. the solution is obviosly wrong. it takes a rock much more than 6.8 sec to fall down from a 227 m high building. you are all neglecting the fact that it's not in vacuum. the formula is g = x" - fx' where f is related to friction. obviously, if f or h are small enough you simply integrate without -fx' from 0 to h and get the high school formula. but h is not small so the exponential term kicks in and the rock travels at the terminal velocity (which is constant) after falling long enough. without knowing the friction you can't do much. you might get the solution if you drop a rock form the edge of the building, then climb on a ladder of known hight and throw it again a few times (once should be enough i think). mesure the time but change the integral boundries when you use the ladders. hey, why didnt the question give the drag constant for the object, density of the medium, or anything else relating to drag? oh right, that's cause it doesnt care about drag! | ||
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