2/15-pr-03
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2/15-pr-03 | | | | | |
| PROBLEMA III, PROPOSITIO XV. | PROBLEM III, PROPOSITION XV |
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| Dato perpendiculo et plano ad ipsum inflexo, partem in perpendiculo infra extenso reperire, quae tempore eodem conficiatur ac planum {30} inflexum post casum ex dato perpendiculo. | Given a vertical line and a plane inclined to it, it is required to find a length on the vertical line below its point of intersection which will be traversed in the same time as the inclined plane, each of these motions having been preceded by a fall through the given vertical line. |
| Sit perpendiculum AB, et planum ad ipsum inflexum BC: oportet, in perpendiculo infra extenso partem reperire, quae ex casu ab A conficiatur {233} tempore eodem atque BC ex eodem casu ab A. Ducatur horizontalis AD, cui occurrat CB extensa in D, et ipsarum CD, DB media sit DE, et BF ponatur aequalis BE; deinde ipsarum BA, AF tertia proportionalis sit AG: dico, BG esse spatium quod post casum AB conficitur tempore eodem ac planum BC post eundem casum. Si enim ponamus, tempus per AB esse ut AB, erit tempus per DB ut DB; et quia DE est media inter BD, DC, erit eadem DE {10} tempus per totam DC, et BE tempus per reliquam BC ex quiete in D, seu ex casu AB. Et similiter concludetur, BF esse tempus per BG, post casum eundem; est autem BF aequalis BE; ergo patet propositum. | Let AB represent the vertical line and BC the inclined plane; it is required to find a length on the perpendicular below its point of intersection, which after a fall from A will be traversed in the same time which is needed for BC after an identical fall from A. Draw the horizontal AD, intersecting the prolongation of CB at D; let DE be a mean proportional between CD and DB; lay {233} off BF equal to BE; also let AG be a third proportional to BA and AF. Then, I say, BG is the distance which a body, after falling through AB, will traverse in the same time which is needed for the plane BC after the same preliminary fall. For if we assume that the time of fall along AB is represented by AB, (Condition 2/03-th-03-cor) then the time for DB will be represented by DB. And since DE is a mean proportional between BD and DC, this same (Condition 2/02-th-02-cor2) DE will represent the time of descent along the entire distance DC (Condition 2/11-th-11) while BE will represent the time required for the difference of these paths, namely, BC, provided in each case the fall is from rest at D or at A. (Condition 2/02-th-02-cor2) (Condition 2/11-th-11) In like manner we may infer that BF represents the time of descent through the distance BG after the same preliminary fall; but BF is equal to BE. Hence the problem is solved. |
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2/15-pr-03 | | | | | |