Physics 114B- MechanicsLecture 9 (Walker: 4.3-5)2D Kinematics ExamplesJanuary 18, 2011Physics 114B- MechanicsLecture 9 (Walker: 4.3-5)2D Kinematics ExamplesJanuary 18, 2011John G. CramerProfessor Emeritus, Department of PhysicsB451 PABjcramer@uw.eduJanuary 18, 2011 Physics 114A- Lecture 9Finding Position from Velocity Finding Position from Velocity2/38January 18, 2011 Physics 114A- Lecture 9Solution:The net distance traveledis the area under the velocitycurve shown in blue.This isa triangle with sides 12 m/sand 3.0 s.The area of thistriangle is:A = (12 m/s)(3 s) = 18 m.Thus, the drag racer moves18 m in the first 3 seconds.Drag Racers Displacement Drag Racers DisplacementThe figure shows thevelocity of a drag racer.How far does the racermove during the first 3.0 s?3/35January 18, 2011 Physics 114A- Lecture 9Upon graduation, a joyful student throws her cap straight up in the air with an initial speed of 14.7 m/s.Given that its acceleration has a magnitude of 9.81 m/s2andis directed downward (we neglect air resistance),(a) When doesthe cap to reach its highest point?(b) What is the distance to the highest point?(c)Assuming the cap is caught at the same height it was released, what is the total time that the cap is in flight?Example: The Flying Cap Example: The Flying Cap1. Draw the cap (as a dot) in its various positions.2. (a) Use the time, velocity and acceleration relation.(b)Use average velocity:vav= v0/2 = 7.35 m/s;Ay = vav At = (7.35 m/s)(1.5 s) = 11.0 m(c)Up time = down time, so total time is 3.0 s.(see text for a more complicated method.)3.The answers have the right units and seem reasonable.002(0 m/s) (14.7 m/s);1.5 s9.81 m/sy yy y yyv vv v a t ta

= + A A= = =

4/35January 18, 2011 Physics 114A- Lecture 9Example: The Flying Cap (continued)Example: The Flying Cap (continued)The height of the cap vs. time has the form of a parabola (since x ~ t2).It is symmetric about the midpoint (but would not be if air resistance were present).The velocity of the cap vs. time has the form of a straight line (since v ~ t).The velocity crosses zero at the midpoint and is negative thereafter, because the cap is moving downward.5/35January 18, 2011 Physics 114A- Lecture 9Zero Launch Angle Zero Launch AngleLaunch angle: direction of initial velocity with respect to horizontalQuestion: Which diver hits the water with the greatest speed?2 2020( cos )( sin ) 2wv vv ghUU= + + |2 202wv v gh = +6/35January 18, 2011 Physics 114A- Lecture 9Zero Launch Angle Zero Launch AngleIn the zero launch angle case, the initial velocity in the y-direction is zero.Here are the equations of motion, with x0= 0 and y0= h:7/35January 18, 2011 Physics 114A- Lecture 9Zero-Launch Trajectory Zero-Launch TrajectoryThis is the trajectory of a projectile launched horizontally.It is a parabola.8/35January 18, 2011 Physics 114A- Lecture 9Zero Launch Angle Zero Launch AngleEliminating t and solving for y as a function of x:This has the form y = a + bx2, which is the equation of a parabola.The landing point can be found by setting y = 0 and solving for x:9/35January 18, 2011 Physics 114A- Lecture 9Projectile Motion Projectile Motion00 0 0 0 0 0 00cos ; sin ; tanyx yxvv v v vvU U U = = =0;x ya a g = = 0 0;x x y yv v v v gt = = 0 0( ) ;xxt x v t = +1 20 02( )yyt y v t gt = + 20 1 202 20 0 0 0 0; ( )2yyx x x x xvx x x gt y x v g x xv v v v v = == ' ' ' ' ' ' ' ' )202 20 0( ) tan2 cosgy x x xvUU = ' 'This is the equationfor a parabola.g10/35January 18, 2011 Physics 114A- Lecture 9General Launch Angle General Launch AngleIn general, v0x= v0cos and v0y= v0sin This gives the equations of motion:11/35January 18, 2011 Physics 114A- Lecture 9Clicker Question 1 Clicker Question 1You throw a ball into the air with an initial speed of 10 m/s at an angle of 60 above the horizontal.The ball returns to the height at which it was thrown in a time T.Which of these plots best represents the speed of the ball as a function of time?12/35January 18, 2011 Physics 114A- Lecture 9General Launch Angle General Launch AngleSnapshots of a trajectory; red dotsare at t= 1 s, t= 2 s, and t= 3 sRange13/35January 18, 2011 Physics 114A- Lecture 9Projectile Motion: Range Projectile Motion: RangeRange: the horizontal distance a projectile travelsIf the initial and final elevation are the same:14/35January 18, 2011 Physics 114A- Lecture 9Projectile Motion:Maximum RangeProjectile Motion:Maximum RangeThe range is a maximum when = 45:15/35January 18, 2011 Physics 114A- Lecture 9Symmetry in projectile motion:Projectile Motion:SymmetryProjectile Motion:SymmetrySame y and |v|Same y and |v|16/35January 18, 2011 Physics 114A- Lecture 9Path of Projectile Path of Projectilexy )202 20 0( ) tan2 cosgy x x xvUU = ' '17/35January 18, 2011 Physics 114A- Lecture 9A delighted physics graduate throws her cap into the air with an initial velocity of 24.5 m/s at a 36.9 angle with the horizontal.The cap is later caught by another student. [Neglect air resistance ;Sin(36.9) = 3/5;Cos(36.9) = 4/5]. (a) Find the total time the cap is in the air.(b) Find the total horizontal distance traveled.Example: A Cap in the Air Example: A Cap in the Air1 202( )yyt v t gt = 10 1 2 02When0, ( ) 0, so t =0 and t =2 /y yy t v gt v g ==0 0 0sinyv v U =22 0 0=2 sin / 2(24.5 m/s)(sin36.9 ) /(9.81 m/s ) 3.00 s t v g U = =0 2 0 0 2cos (24.5 m/s)(cos36.9 )(3.00 s) 58.8 mxx v t v t U = = = =g18/35January 18, 2011 Physics 114A- Lecture 91 202d dgr v t gt = +& & & 1 202m mr r gt = +& & &r = 00Therefore, when (i.e., the dart hits the monkey) thenm d mo dgr r r v t = =& & & &In other words,provided the gun is aimed so that the extension of thevector vdg0 passes through the monkeys position rm0, the dart hits the monkey.A park ranger with a tranquilizer dart intends to shoot amonkey hanging from a branch.The ranger points the barreldirectly at the monkey, not realizing that the dart willfollow a parabolic path that will pass below themonkeys present position.The monkey,seeing the gun discharge, instantly lets go of the branch and drops out of thetree, expecting to avoid the dart.(a) Show that the monkey will be hit,independent of the dart velocity vd.(b) Let vd0be the initial velocity of the dart.Find the dart velocity relative to the monkey at an arbitrary time t during the darts flight.Example: Ranger & Monkey Example: Ranger & Monkey19/35January 18, 2011 Physics 114A- Lecture 9A helicopter drops a supply package to flood victims on a raft in a swollen lake.When the package is released, the helicopter is 100 m directly above the raft and flying at a velocity of 25.0 m/s at an angle of 36.9 above the horizontal. [Neglect air resistance;Sin(36.9) = 3/5;Cos(36.9) = 4/5].(a) How long is the package in the air?(b) How far from the raft does the package land?(c)If the helicopter continues at constant speed, where is it when the package lands?Example: A Supply Drop Example: A Supply Drop1 202( )yyt v t gt = 1 2020ygt v t y =+0 0 0 0 0 0sin (25.0 m/s)(sin36.9 ) 15.0 m/s;cos 20.0 m/sy xv v v v U U = = = = =22 20 022(15.0 m/s) (15.0 m/s) 2(9.81 m/s )( 100 m)(9.81 m/s )y yv v gytgs s= =3.24 s and6.30 s t t ==0(20.0 m/s)(6.30 s) 126 mh xx x v t = = = =0 0 00 (15.0 m/s)(6.30 s) 94.4 mh h h yy y v t v t = + = + = =The helicopter will be 194 m above the package impact point.g20/35January 18, 2011 Physics 114A- Lecture 9Example: A Rocket Sled Example: A Rocket SledA rocket sled acceleratesat 50 m/s2for 5.0 s, coastsfor 3.0 s, then deploys abraking parachute anddecelerates at 3.0 m/s2until coming to a halt.(a) What is the maximumvelocity reached bythe rocket sled?v1x = v0x+ a0x(t1 t0)= a0xt1= (50 m/s2)(5.0 s)= 250 m/s(b) What is the total distance traveled bythe rocket sled?x1 = x0+ v0x(t1 t0) + a0x(t1 t0)2= a0xt12= (50 m/s2)(5.0 s)2= 625 mx2= x1+ v1x(t2 t1) = 625 m + (250 m/s)(3.0 s) = 1375 mv3x2= 0 = v2x2+2a2xAx = v2x2+2a2x(x3 x2), sox3= x2+ (v3x2- v2x2)/2a2x= 1375 m + [0 (250 m/s)2]/[2(-3.0 m/s2)] = 11,800 m21/35January 18, 2011 Physics 114A- Lecture 9A person skateboarding witha constant speed of 1.30 m/sreleases a ball from a heightof 1.25 m above the ground.Use x0 = 0 and y0 = h = 1.25 m.(a)Find x and y fort1= 0.250 s.(b)Find x and y fort2= 0.500 s.(c)Find the velocity, speed, and direction of the ball att2= 0.500 s.Example: Dropping a Ball Example: Dropping a Ball1 0 1(1.30 m/s)(0.250 s) 0.325 m x vt = = =1 1 2 2 21 12 2(1.25 m) (9.81 m/s )(0.250 s) 0.943 m y h gt ===2 0 2(1.30 m/s)(0.500 s) 0.650 m x vt = = =1 1 2 2 22 22 2(1.25 m) (9.81 m/s )(0.500 s) 0.0238 m y h gt ===01.30 m/sxv v = =22(9.81 m/s )(0.500 s) 4.91 m/syv gt === (1.30 m/s) ( 4.91 m/s) v x y = +&2 2 2 2(1.30 m/s) ( 4.91 m/s) 5.08 m/sx yv v v = + = + = ) . Jarctan / arctan( 4.91 m/s) / (1.30 m/s) 75.2y xv v U= ==22/35January 18, 2011 Physics 114A- Lecture 9From the same height (and at the same time), one ball is dropped and another ball is fired horizontally.Which one will hit the ground first?Clicker Question 2:Dropping the BallClicker Question 2:Dropping the Ball(A)the dropped ball (A)the dropped ball(B)the fired ball (B)the fired ball(C)they both hit at the same time (C)they both hit at the same time(D)it depends on how hard the ball was fired (D)it depends on how hard the ball was fired(E)it depends on the initial height (E)it depends on the initial height23/35January 18, 2011 Physics 114A- Lecture 9Chipping from the rough, a golfer sends the ball over a 3.0 m high tree that is 14.0 m away. The ball lands on the green at the same level from which it was struck after traveling a horizontal distance of 17.8 m.Example: A Rough Shot Example: A Rough Shot(a)If the ball left the clubat 54.0 above the horizontaland landed 2.24 s later, what was its initial speed v0?(b)How high was the ball when it passed over the tree?/ (17.8 m) / (2.24 s) 7.95 m/sxv d t = = =0/ cos (7.95 m/s) / cos54.0 13.5 m/sxv v U = = =1/ (14.0 m) / (7.95 m/s) 1.76 sxt x v = = =1 1 2 20 1 1 0 1 12 21 2 22sin(13.5 m/s)sin54.0(1.76 s) (9.81 m/s )(1.76 s) 4.03 myy y vt gt v t gt U = += = =t124/35January 18, 2011 Physics 114A- Lecture 9A baseball is hit so that it leaves the bat making a 30 angle with the ground.It crosses a low fence at the boundary of the ballpark 100 m from home plate at the same height that it was struck.(Neglect air resistance.)What was its velocity as it left the bat?Example: A Home Run Example: A Home Run0 0 0 01 0 0 1 0 0 11 1 2 21 0 0 1 0 1 0 0 1 12 2cos ; sin ;( ) ( cos ) ;0 ( ) ( ) ( sin ) ;x yxyv v v vx x v t t v ty y v t t gt t v t gtU UUU= == +== = += 1 1 20 1 1 0 1 12 21 1 00 ( sin ) ( sin ) ;so0 or2 sin / ;v t gt v gt tt t v gU UU== = =1 0 1 0 02 21 0 020 1Therefore,( cos ) ( cos )(2 sin / );2 sin cos / sin 2 / ;/ sin 2 (100 m)(9.80 m/s ) / sin(60 ) 33.6 m/sx v t v v gx v g v gv x gU U UU U UU= == == = =g25/35January 18, 2011 Physics 114A- Lecture 9Clicker Question 3:War at SeaClicker Question 3:War at SeaA battleship simultaneously fires two shells at two enemyA battleship simultaneously fires two shells at two enemy submarines.The shells are launched with the same initial velocity. submarines.The shells are launched with the same initial velocity.If the shells follow the trajectories shown, which submarine getsIf the shells follow the trajectories shown, which submarine gets hit first ? hit first ?1 2(A)Submarine 1 (A)Submarine 1(B)Submarine 2 (B)Submarine 2(C)They are both hit at the same time (C)They are both hit at the same time(D)It depends on the initial velocity (D)It depends on the initial velocity26/35January 18, 2011 Physics 114A- Lecture 9Horizontal Range of a Projectile Horizontal Range of a Projectile1 2020 ; 0yv T gT T =>1020yv gT=00022sinyvvTg gU = = )20 00 0 0 0 0 02cos sin 2sin cosxv vR v T vg gU U U U = = = ' '0 0 02sin cos sin 2 U U U =200sin 2vRgU = is maximum when Uo=45o, so that sin 2Uo = 1.g27/35January 18, 2011 Physics 114A- Lecture 9A golfer hits a ball from the origin with an initial speed of 30.0 m/s at an angle of 50.0 above the horizontal.The ball lands on a green that is 5.00 m above thelevel where the ball was struck.Example: An Elevated Green Example: An Elevated Green(a)How long was the ball in the air?(b) How far did the ball travel horizontally before ot landed?(c) What is the speed and direction of the ball just before it lands?1 1 1 2 2 20 1 0 12 2 2sin 0y yy y vt gt v t gt h h vt gt U = +== + = )22 / 2 0.229 s and4.46 sy yt v v hg h t t = s = =0cos (30.0 m/s) cos50(4.46 s) 86.0 mxx vt v t U = = = =0cos (30.0 m/s) cos50 19.3 m/sxv v U = = =20sin (30.0 m/s) cos50 (9.81 m/s )(4.46 s) 20.8 m/syv v gt U == = 2 228.4 m/sx yv v v = + =. Jarctan / arctan( 20.8 m/s) / (19.3 m/s) 47.1y xv v U = ==28/35January 18, 2011 Physics 114A- Lecture 9A police officer chases a master jewelthief across city rooftops.They are bothrunning when they come to a gap betweenbuildings that is 4.0 m wide and has a dropof 3.0 m.The thief having studied a littlephysics, leaps at 5.0 m/s at an angle of 45 above the horizontal and clears the gap easily.The police officer did not study physics and thinks he should maximize his horizontal velocity, so he leaps horizontally at 5.0 m/s.(a) Does he clear the gap?(b) By how much does the thief clear the gap?Example: To Catch a Thief Example: To Catch a Thief1 20 02yy y v t gt = + 1 2 223.0 m 0 0 (9.81 m/s )t= + 26.0 m/ 9.81 m/s 0.782 s t = =0 00 (5.0 m/s)(0.782 s) 3.91 mxx x v t = + = + =1 2 223.0 m 0 (5.0 m/s) sin 45 (9.81 m/s ) t t= + 0.50 sor 1.22 s t t ==0 00 (5.0 m/s)(1.22 s) cos 45 4.31 mxx x v t = + = + =4.31 m 4.00 m 0.31 m x A ==No!29/35January 18, 2011 Physics 114A- Lecture 9The archerfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fishs mouth.Suppose an archerfish squirts water with an initial speed of 2.30 m/s at an angle of 19.5 above the horizontal.When the water reaches a beetle ata height h above the waterssurface, it is moving horizontally.Example: What A Shot! Example: What A Shot!(a)How much time does the beetle have to react?(b) What is the height h of the beetle?(c) What is the horizontal distance d between the fish and the beetle?0 0sin 0y yv v gt v gt U ===20sin / (2.30 m/s)sin19.5 / (9.81 m/s ) 0.0782 s t v g U = = =1 1 2 20 02 21 2 22sin(2.30 m/s) sin19.5(0.0782 s) (9.81 m/s )(0.0782 s) 0.0300 myh y vt gt v t gt U = += = = =0 0cos (2.30 m/s) cos19.5(0.0782 s) 0.170 mxd x vt v t U = + = = =30/35January 18, 2011 Physics 114A- Lecture 9Example: Basketball Throw Example: Basketball ThrowA basketball player stands 15.0 ft from the basket, which is 10.0 ft above the floor.The ball leaves his hands 6.0 ft above the floor with an initial velocity v0that is directed at an angle U with the horizontal.What are the conditions on v0and U such that he makes the basket?00coscosffxx v t tvUU= = )1 2022020022 20sinsincos2 costan2 cosf ifi ffi fy y v t g tg xvy xvvg xy xvUUUUUU= + = + = + 0cos 2( tan )ff i fxgvx y y U U=+ 0 20 40 60 80101520v0Uv0= 7.64 m/sU = 52.531/35January 18, 2011 Physics 114A- Lecture 9Summary - 2D Kinematics Summary - 2D Kinematics Components of motion in the x- and y-directions can be treated independently. In projectile motion, the acceleration is g. If the launch angle is zero, the initial velocity has only an x-component. The path followed by a projectile is a parabola. The range is the horizontal distance the projectile travels.32/35