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 [email protected] 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
