Projectile Motion. What is projectile? Projectile -Any object which projected by some means and continues to move due to its own inertia (mass).

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1 Prjectile Mtin AP Phyic B What i prjectile? Prjectile -Any bject which prjected by me mean and cntinue t me due t it wn inertia (ma). 1

2 Prjectile me in TWO dimenin Since a prjectile me in - dimenin, it therefre ha cmpnent jut like a reultant ectr. n Hrizntal and Vertical Hrizntal Velcity Cmpnent n NEVER change, cer equal diplacement in equal time perid. Thi mean the initial hrizntal elcity equal the final hrizntal elcity In ther wrd, the hrizntal elcity i CONSTANT. BUT WHY? Graity DOES NOT wrk hrizntally t increae r decreae the elcity.

3 Vertical Velcity Cmpnent n Change (due t graity), de NOT cer equal diplacement in equal time perid. Bth the MAGNITUDE and DIRECTION change. A the prjectile me up the MAGNITUDE DECREASES and it directin i UPWARD. A it me dwn the MAGNITUDE INCREASES and the directin i DOWNWARD. Cmbining the Cmpnent Tgether, thee cmpnent prduce what i called a trajectry r path. Thi path i parablic in nature. Cmpnent Magnitude Directin Hrizntal Cntant Cntant Vertical Change Change 3

4 Hrizntally Launched Prjectile Prjectile which hae NO upward trajectry and NO initial VERTICAL elcity. = = cntant x y = 0 m/ Hrizntally Launched Prjectile T analyze a prjectile in dimenin we need equatin. One fr the x directin and ne fr the y directin. And fr thi we ue kinematic #. x = 1 t + at x = t y = 1 gt Remember, the elcity i CONSTANT hrizntally, that mean the acceleratin i ZERO! Remember that ince the prjectile i launched hrizntally, the INITIAL VERTICAL VELOCITY i equal t ZERO. 4

5 Hrizntally Launched Prjectile Example: A plane traeling with a hrizntal elcity f 100 m/ i 500 m abe the grund. At me pint the pilt decide t drp me upplie t deignated target belw. (a) Hw lng i the drp in the air? (b) Hw far away frm pint where it wa launched will it land? y= 1 gt 500 = 1 ( 9.8) t = t t = 10.1 ecnd What d I knw? =100 m/ t =? y = 500 m x =? y = 0 m/ g = -9.8 m// What I want t knw? x= t = (100)(10.1) = 1010 m Vertically Launched Prjectile NO Vertical Velcity at the tp f the trajectry. Vertical Velcity decreae n the way upward Hrizntal Velcity i cntant Vertical Velcity increae n the way dwn, Cmpnent Magnitude Directin Hrizntal Cntant Cntant Vertical Decreae up, tp, Increae dwn Change 5

6 Vertically Launched Prjectile Since the prjectile wa launched at a angle, the elcity MUST be brken int cmpnent!!! y y = = cθ inθ θ Vertically Launched Prjectile There are eeral thing yu mut cnider when ding thee type f prjectile beide uing cmpnent. If it begin and end at grund leel, the y diplacement i ZERO: y = 0 6

7 Vertically Launched Prjectile Yu will till ue kinematic #, but YOU MUST ue COMPONENTS in the equatin. y x 1 = t y = yt + gt θ y = = cθ inθ Example A place kicker kick a ftball with a elcity f 0.0 m/ and at an angle f 53 degree. (a) Hw lng i the ball in the air? (b) Hw far away de it land? (c) Hw high de it trael? θ = 53 y y = cθ = 0c 53 = 1.04 m/ = inθ = 0in 53 = m/ 7

8 Example A place kicker kick a ftball with a elcity f 0.0 m/ and at an angle f 53 degree. (a) Hw lng i the ball in the air? What I knw =1.04 m/ t =? y =15.97 m/ x =? What I want t knw y = 0 y max =? g = m// y = 1 yt+ gt 0 = (15.97) t 4.9t 15.97t = 4.9t = 4.9t t = 3.6 Example A place kicker kick a ftball with a elcity f 0.0 m/ and at an angle f 53 degree. (b) Hw far away de it land? What I knw =1.04 m/ What I want t knw t = 3.6 y =15.97 m/ x =? y = 0 y max =? g = m// x= t (1.04)(3.6) = 39.4 m 8

9 Example A place kicker kick a ftball with a elcity f 0.0 m/ and at an angle f 53 degree. (c) Hw high de it trael? CUT YOUR TIME IN HALF! What I knw =1.04 m/ y =15.97 m/ What I want t knw t = 3.6 x = 39.4 m y = 0 y max =? g = m// 1 y = yt+ gt y = (15.97)(1.63) 4.9(1.63) y = m 9

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