As we have already discussed, all the objects have the same absolute value of

Transcription

1 Lecture 3 Prjectile Mtin Lst time we were tlkin but tw-dimensinl mtin nd intrduced ll imprtnt chrcteristics f this mtin, such s psitin, displcement, elcit nd ccelertin Nw let us see hw ll these thins re interrelted in the specil cse f cnstnt ccelertin We lred knw tht in this cse equtins f mtin will he the frm cnst, cnst, cnst t, t, t 1 r r t t, 1 t t, 1 t t (31) (3) (33) As we he lred discussed, ll the bjects he the sme bslute lue f ccelertin, 9 8m s, when the re fllin dwn ner the erth s surfce nd this ccelertin is pintin dwnwrds t the erth's surfce T simplif the prblem we shll inre ir resistnce, which is quite ccurte pprimtin s fr s speed f the bject is nt t lre nd there is n wind We shll nl cnsider the prblem in tw dimensins S, we will nl stud the bject which mes in erticl plne We chse the -is t be hrizntl is in the sme plne s the bject mes nd is t be erticl is in the sme plne with psitie directin in upwrds The bject hs n riinl elcit,, in the sme plne (see the picture) Such n bject is clled prjectile nd its mtin prjectile mtin

2 We cn resled the riinl elcit int tw cmpnents which is i j, cs, sin (34) Here,, is the nle between the directin f is nd the riinl elcit We cn then resle the ccelertin ectr, but it is een esier The nl cmpnent tht ccelertin hs is erticl cmpnent,, (35) j Since rittinl ccelertin is cnstnt we cn cnsider this prblem usin the set f equtins It cn be reduced t tw ne-dimensinl prblems, becuse fr prjectile erticl mtin nd hrizntl mtin re cmpletel independent This fct cn be cnfirmed eperimentll b wtchin tw blls, ne f which is fllin striht dwn, nther is sht hrizntll b sprin Their mtin in erticl directin will be ectl the sme The will cer the sme distnces durin the sme time interls nd it tkes the sme time fr bth f them t rech the rund

3 Fr hrizntl mtin we he, cs cnst, t cs t Fr erticl mtin, the similr set f equtins ies cnst, t sin t, t t t sin t (36) (37) The lst set f equtins illustrtes behir f the erticl cmpnent f elcit depicted in m picture At first it is equl then it ets smller due t decelertin f the bject b rittinl field At the hihest pint f the trjectr the erticl cmpnent f elcit becmes zer, s it nl mes hrizntll, then it is ccelerted in the dwnwrd directin b the sme rittinl field Finll it reches the rund with the sme speed s it hd riinll, nd it mkes the sme nle belw the hrizn s the riinl nle be the hrizn ws The trjectr m lk like cmplicted cure Hweer, it cn be shwn b elimintin time frm equtins 36 nd 37, tht it is just prbl tn, (38) (fr simplicit, we set ( cs ), ) Eercise: Pre equtin 38 Eercise: Hw equtin 38 will lk like if nd? In the picture u cn ls see the hrizntl distnce, R, which prjectile hs treled befre it returns t its initil lunchin leel Let us find this distnce, clled the hrizntl rne f the prjectile Fr the crdintes t the finl pint ne hs R, With ccunt f equtins 36, 37 it becmes

4 R cs t, t sin t Elimintin time frm these tw equtins ne hs R sin cs sin (39) Eercise: Pre equtin 39, b elimintin time frm the preius equtin This lst equtin cn help us t nswer the interestin questin: At which nle ne needs t sht frm the un, in rder t et the lrest hrizntl rne R? If R hs the lrest lue fr the ien riinl speed f prjectile, this mens tht sine f the nle in equtin 39, shuld rech its mimum lue, which is 1, s we he sin 1, 9 45, This mens tht hrizntl rne will rech its mimum, if u lunch prjectile t 45 derees nle nd it is equl R m sin 9 Emple 31 At wht nle reltie t the hrizn ne hs t sht prjectile in rder t he the prjectile's heiht t the hihest pint equl prjectile's hrizntl rne? We just pred tht prjectile's hrizntl rne is R sin cs Nw let us see wht prjectile's heiht is Since t the hihest pint f the trjectr the erticl cmpnent f elcit is zer, we he ( ), where, sin, nd H, s (sin ) H, H (sin ), (sin ) H

5 In this prblem we he H=R, s (sin ) sin cs, sin 4, cs tn 4, 1 tn (4) 76 sin cs,