1 cos. where v v sin. Range Equations: for an object that lands at the same height at which it starts. v sin 2 i. t g. and. sin g

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1 SPH3UW Unt.5 Projectle Moton Pae 1 of 10 Note Phc Inventor Parabolc Moton curved oton n the hape of a parabola. In the drecton, the equaton of oton ha a t ter Projectle Moton the parabolc oton of an object, where the horzontal coponent rean contant and the vertcal oton chane wth te. Rane the dtance a projectle travel n the horzontal drecton. Intal Veloct Anle the anle of a projectle oton, eaured relatve to the horzontal unle otherwe tated. Dtance Equaton: drecton: d v t where v v 1 co drecton: d v t 4.9 t where v v n Rane Equaton: for an object that land at the ae heht at whch t tart v n d and v n t Mau Heht: for an object that land at the ae heht at whch t tart v n h and t v n The horzontal rane R a au for a launch anle of 45. Projectle oton the cobnaton of two tpe of kneatc proble: one nvolvn acceleraton due to ravt, and the other nvolvn a contant veloct n one drecton. In projectle oton, the horzontal oton and the vertcal oton are ndependent of each other; that, nether oton affect the other. Note: We have aued that the ar throuh whch the projectle ove ha no effect on t oton. However, n an tuaton, the dareeent between our calculaton and the actual oton of the projectle can be lare becaue the ar ret the oton.

2 SPH3UW Unt.5 Projectle Moton Pae of 10 Step for Solvn Projectle Moton Proble 1. Denate a n conventon for drecton. Mot proble ue + for oton that ether upward or to the rht. In cae where the object alwa falln, ou a fnd t advantaeou to denate + for downward oton.. Separate the ven value nto ther horzontal () and vertcal coponent () 3. Wrte down the correpondn equaton for each coponent. 4. If ou can calculate TIME n ether the or drecton then do o, and the ue our anwer to calculate the rane or heht. 5. If the equaton n the and drecton each have two unknown value then ue the ubttuton ethod to olve. Note: The horzontal ntal veloct ean that v 0. If the projectle land at the ae heht t tarted fro, then d 0. Once the object leave the table, t eperence a downward acceleraton equal to ravt. Thu the vertcal veloct( v ) contnuall ncrean. The horzontal veloct( v ) rean contant and equal to v. The two vector v and v are added toether to et the veloct at each pont on the path. If an object ponted at an anle, the oton eentall the ae ecept that there now an ntal vertcal veloct ( v ). Becaue of the downward acceleraton of ravt, v contnuall decreae untl t reache t hhet pont, at whch t ben to fall downward.

3 SPH3UW Unt.5 Projectle Moton Pae 3 of 10 Eaple: A baeball thrown wth an ntal horzontal veloct of 85 / at an anle of 50 above the horzontal. How far doe the baeball travel? Soluton: We are ven v 85, 50, d? We wll denate the potve drecton a to the rht and upward. Now d v t v co t 85 co50 t t We need t, whch wll coe fro the horzontal coponent to the flht. Un d v t 4.9t v n t 4.9 t 85 n t 4.9 t t t Solvn wth the quadratc forula t b b 4ac a Subttutn th nto d t nfcant dt

4 SPH3UW Unt.5 Projectle Moton Pae 4 of 10 Eaple A ball toe fro the top of a cence buldn wth a horzontal veloct coponent of 15 / [rht] and a vertcal coponent of 0 / [up]. If the roof wa 100 hh, how far wll the ball travel? Soluton: We are ven: v 15, v 0 Let frt fnd the te for the trp fro the horzontal coponent. d v t 4.9 t t 4.9 t 4.9 t 0t100 0 d Now un the quadratc forula to olve for t.

5 SPH3UW Unt.5 Projectle Moton Pae 5 of 10 t b b 4ac a or.9 Snce a neatve te poble n real lfe, the correct te 7.0. To deterne the rane, we ubttute th value nto The ball travel 105 d v t Eaple A ball roll of a lanted roof (lant 30 wth repect to the horzontal). The ball rolln wth a peed of 4.1 / fall 6.0 etre to the round. a) how lon wa the ball n the ar after t left the roof? b) at what wa the horzontal dtance fro the roof dd the ball ht the round? c) What wa the pact veloct of the ball? Soluton: a) Fro d v t 4.9t 6 4.1n 30 t 4.9 t 4.9 t.05 t 6 0 Fro the quadratc forula

6 SPH3UW Unt.5 Projectle Moton Pae 6 of 10 t b b 4ac a or Therefore the ball took 0.9 econd to ht the round b) Fro d v t 4.1 co c) We wll chooe down to be potve Fro v v a t f v co 30

7 SPH3UW Unt.5 Projectle Moton Pae 7 of 10 Balltc Equaton The follown equaton are ueful for calculatn Rane, Mau heht and te n ar onl for a etrc path, a hown below. If the path not etrc, ou ut ue the prevou acceleraton forula. Un v f v at for vertcal drecton we obtan to te for an object that launche at the ae alttude to arrve back down at the ae alttude [rane te]: v n v n t v n v v at f t v n t Therefore, t take half of th te to reach the au heht: t v n We can ubttute th value nto 1 vt at to deterne forula for au heht. vn 1 vn h v n v n

8 SPH3UW Unt.5 Projectle Moton Pae 8 of 10 For rane, we ue: d vt v n co v n v co v n Rane Equaton: for an object that land at the ae heht at whch t tart v n v n d and t Mau Heht: for an object that land at the ae heht at whch t tart v n h and t v n The horzontal rane R a au for a launch anle of 45. Rane eaple A lon-juper leave the round at an anle of 0.0 above the horzontal and at a peed of 11.0 /. a) How far doe he jup n the horzontal drecton? (Aue no ar retance). b) What the au heht reached? c) How lon wa he n the ar? Soluton: Snce the juper launche and land at the ae reference heht, we can ue our balltc equaton to plf the oluton. a) Rane: d v n 11.0 n

9 SPH3UW Unt.5 Projectle Moton Pae 9 of 10 b) Heht: h v n 11.0 n c) Te n ar: v n t 11.0 n

10 SPH3UW Unt.5 Projectle Moton Pae 10 of 10 Etra Note and Coent