Zanotti, Francesco Maria, Della forza de' corpi che chiamano viva libri tre, 1752

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              <pb o="47" file="0071" n="71" rhead="LIBRO I."/>
            po A
              <emph style="it">b</emph>
            ; </s>
            <s xml:id="echoid-s714" xml:space="preserve">che ben ſapete, lo ſpazio, che un corpo
              <lb/>
            ſcorre, eſsere la velocità moltiplicata per lo tem-
              <lb/>
            po. </s>
            <s xml:id="echoid-s715" xml:space="preserve">Così è, diſse il Signor Marcheſe, poichè eſ-
              <lb/>
            ſendo
              <emph style="it">s</emph>
            lo ſpazio, il tempo
              <emph style="it">t</emph>
            , la velocità ſarà {
              <emph style="it">s/t</emph>
            }
              <lb/>
            che moltiplicata per
              <emph style="it">t</emph>
            rende
              <emph style="it">s</emph>
            . </s>
            <s xml:id="echoid-s716" xml:space="preserve">E per ciò, ripigliai
              <lb/>
            io, il rettangoletto
              <emph style="it">br</emph>
            , che pur ſi fa moltiplican-
              <lb/>
            do la velocità A
              <emph style="it">r</emph>
            per lo tempetto A
              <emph style="it">b</emph>
            , eſprime-
              <lb/>
            rà lo ſpazio ſcorſo in eſso tempetto A
              <emph style="it">b</emph>
            - Vedete
              <lb/>
            dunque, che come il corpo ſarà caduto per lo
              <lb/>
            piccoliſsimo tempo A
              <emph style="it">b</emph>
            , la velocità, che egli av-
              <lb/>
            rà, ſarà
              <emph style="it">bc</emph>
            eguale ad A
              <emph style="it">r</emph>
            , e lo ſpazio fcorſo ſarà
              <lb/>
            il rettangoletto
              <emph style="it">br</emph>
            . </s>
            <s xml:id="echoid-s717" xml:space="preserve">Ma, ſcorſo lo ſpazio
              <emph style="it">br</emph>
            , ri-
              <lb/>
            ceverà il corpo ſul principio del tempetto
              <emph style="it">bd</emph>
            un’
              <lb/>
            altro impulſo dalla gravità eguale a quel primo,
              <lb/>
            laonde ritenendo la velocità
              <emph style="it">bc</emph>
            , che già avea, ne
              <lb/>
            acquiſterà un’ altra
              <emph style="it">ct</emph>
            ad eſsa eguale; </s>
            <s xml:id="echoid-s718" xml:space="preserve">e verrà nell’
              <lb/>
            intervallo
              <emph style="it">bd</emph>
            a ſcorrere con la velocità
              <emph style="it">bt</emph>
            un’ al-
              <lb/>
            tro ſpazietto, che ſarà il rettangolo
              <emph style="it">dt</emph>
            . </s>
            <s xml:id="echoid-s719" xml:space="preserve">E qui
              <lb/>
            pur vedete, che eſsendo il corpo caduto per lo
              <lb/>
            piccioliſſimo tempo A
              <emph style="it">d</emph>
            , la velocità, che egli
              <lb/>
            avrà, ſarà
              <emph style="it">de</emph>
            eguale a
              <emph style="it">bt</emph>
            ; </s>
            <s xml:id="echoid-s720" xml:space="preserve">e lo ſpazio ſcorſo
              <lb/>
            ſarà la ſomma de due rettangoli
              <emph style="it">br</emph>
            ,
              <emph style="it">dt</emph>
            . </s>
            <s xml:id="echoid-s721" xml:space="preserve">E ſe
              <lb/>
            all’ iſteſso modo proſeguirete, faccendo a cia-
              <lb/>
            ſcun tempetto il ſuo rettangolo corriſpondente,
              <lb/>
            facilmente ritroverete, che eſsendo il corpo
              <lb/>
            caduto per qualſiſia aſſegnabil tempo A
              <emph style="it">m</emph>
            , et
              <lb/>
            eſſendo
              <emph style="it">mo</emph>
            il rettangolo corriſpondente all’ ul-
              <lb/>
            timo tempetto, la velocità del corpo ſarà
              <emph style="it">mn</emph>
              <lb/>
            lato del rettangolo
              <emph style="it">mo</emph>
            , e lo ſpazio ſcorſo </s>
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