Physics 111: Mechanics Lecture 2

Bin Chen

NJIT Physics Department Registering your iClicker in class v Smart devices are not accepted. v Turn on your iClicker. v Make sure to use the channel AB. v When you see your name, press the letters shown beside it. v You are registered! v If you make a mistake, press DD to cancel. Let’s test it…

What is the Most Advanced Physics Course You Have Had? A. High school AP Physics course B. High school regular Physics course C. College non-calculus-based course D. College calculus-based course (or I am retaking Phys 111) E. None, or none of the above Announcements q iClicker: please procure your iClicker. We’ll start setting it up today. Don’t lose your quiz credits (10% of your final grade). q PHYS 111 tutoring schedule updated. See our course website https://web.njit.edu/~binchen/phys111/ for detailed information Lecture 1 Review: Problem-Solving Hints

q Read the problem q Draw a diagram n Choose a coordinate system, label initial and final points, indicate a positive direction for velocities and accelerations

q Label all quantities, be sure all the units are consistent n Convert if necessary q Choose the appropriate kinematic equation q Solve for the unknowns n You may have to solve two equations for two unknowns q Check your results Chapter 3 Motion in 2-D or 3-D q Introduction to vectors q 3.1 Position and Velocity Vectors q 3.2 The Acceleration Vector q 3.3 Projectile Motion q 3.4 Motion in a Circle q 3.5* Relative Velocity (self-study section) Vectors and Scalars

q Vectors q Scalars: n Displacement n Distance

n Velocity (magnitude and n Speed (magnitude of direction!) velocity)

n Acceleration n Temperature

n Force n Mass

n Momentum n Energy n Time qA vector quantity has both magnitude (value + unit) and direction qA scalar is completely specified by only a magnitude (value + unit) Vectors: Important Notation q To describe vectors we will use:

n The bold font: Vector A is A Ending position n And/or an arrow above the vector: ! n In the pictures, we will always show vectors as arrows ! n Arrows point the direction

n To describe the magnitude of a vector we will use absolute value

sign: ! or just A Starting position n Magnitude is always positive, the magnitude of a vector is equal to the length of a vector. Properties of Vectors q Equality of Two Vectors

n Two vectors are equal if they have the same magnitude and the same direction q Movement of vectors in a diagram

n Any vector can be moved parallel to itself without being affected q Negative of a vector n A vector the negative of another if they have the same magnitude but are 180apart (opposite directions) ! A ! ! = −$ or $ = −! B Adding Vectors Geometrically (Triangle Method) ! q Draw the first vector A with the appropriate length and in the direction specified, with respect ! ! to a coordinate system A + B ! ! q Draw the next vector B with the B appropriate length and in the direction specified, with respect to a coordinate system whose! origin is the end of vector and A ! parallel to! the coordinate system used for A : tip-to-tail. A q The resultant! is drawn from the origin of! A to the end of the last vector B Adding Vectors Graphically q When you have many vectors, just keep ! ! repeating the process A + B until all are included ! ! ! A+ B+C q The resultant is still drawn from the origin ! ! of the first vector to A + B the end of the last vector More on Vector Addition Vector Subtraction q Special case of vector addition ! B n Add the negative of the subtracted vector ! − # = ! + (−#) ! q Continue with standard A vector addition procedure −# ! − # More on Vector Subtraction

! = ! − " " Components of a Vector q Components of a vector are the projections of the vector along the x- and y-axes q Components are not vectors, they are magnitudes of component vectors

! = !# + !% q

Components of a vector !: 56 = A cos 8 ° 59 = A cos (90 − 8)=A sin 8 Components of a Vector q The previous equations are valid only if θ is measured with respect to the x-axis q The components can be positive or negative and will have the same units as the original vector θ=0, Ax=A>0, Ay=0 θ=45, Ax=Acos45>0, Ay=Asin45>0 ax < 0 ax > 0 θ=90, Ax=0, Ay=A>0 ay > 0 ay > 0 θ θ=135, Ax=Acos135<0, Ay=Asin135>0 ax < 0 ax > 0 θ=180, Ax=-A<0, Ay=0 ay < 0 ay < 0 θ=225, Ax=Acos225<0, Ay=Asin225<0 θ=270, Ax=0, Ay=-A<0 θ=315, Ax=Acos315<0, Ay=Asin315<0 Calculations Using Components: Magnitude and Direction

• We can use the components of a vector to find its magnitude and direction:

% % ! = # = !$ + !'

, , tan + = -, and + = arctan - ,. ,. Calculations Using Components: Vector Addition

• We can use the components of a set of vectors to find the components of their sum: ( = ! + &

• Components of !: "#, "%

• Components of &: '#, '% • Components of (:

)# = "# + '#, )% = "% + '% Unit Vectors

q Unit vectors "̂, #̂, %$ q Unit vectors used to specify direction q Unit vectors have a magnitude of 1

q Then &' = )*"̂, &+ = ),#̂

Magnitude Magnitude

Unit vector Unit vector

& = &' + &+ = )*"̂ + ),#̂ Adding Vectors Algebraically q Consider two vectors

# = '(,̂ + '-.̂

% = )(,̂ + )-.̂ q Then

# + % = ('(,̂ + '-.)̂ + ()(,̂ + )-.)̂

=('(+)(),̂ + ('-+)-).̂ q ! = # + % = ('( + )(),̂ + ('-+)-).̂

! = /(,̂ + /-.̂ q So we have reproduced

0( = '( + )(, 0- = '- + )- Motion in two dimensions q Kinematic variables in one dimension n Position: !(#) m n Velocity: %(#) m/s x n Acceleration: &(#) m/s2 q Kinematic variables in three dimensions

n Position: ' # = )+̂ + -.̂ + /10 m 0 n Velocity: % # = 23+̂ + 24.̂ + 251 m/s 0 2 n Acceleration: & # = 63+̂ + 64.̂ + 651 m/s q All are vectors: have direction and magnitudes Position and Displacement q In one dimension

∆" = "$ %$ − "' %'

x (t ) = - 3.0 m, x (t ) = + 1.0 m ! ! ! i i f f Dr = r2 -r1 Δx = +1.0 m + 3.0 m = +4.0 m q In two dimensions

n Position: the position of an object is described by its position vector ((%)    always points to particle from origin. ri + Δr = rf

n Displacement: ∆( = ($ − (' Example: Walkman q A man walking (with a walkman) firstly takes one walk which can be described algebraically as ! = −3&̂ + 5)̂, followed by another * = 4&̂ − 2)̂. Find the final displacement and direction of the sum of these motions

- = ! + * = (/0 + 10)&̂ + (/3+13))̂ = −3 + 4 &̂ + 5 − 2 )̂ = 1&̂ + 3)̂

50 = 1 53 = 3

2 2 1/ 2 2 2 1/ 2 C = (Cx + Cy ) = (1 + 3 ) = 3.16

53 3 6 = arctan = arctan = 71.5° 50 1 Average Velocity q The average velocity between two points is the displacement divided by the time interval between the two points. q The average velocity has the same direction as the displacement. ∆1 0 = +, ∆)

∆#%̂ + ∆'(̂ ∆# ∆' = = %̂ + (̂ = * %̂ + * (̂ ∆) ∆) ∆) +,,. +,,/ University Physics, 13/e Young/Freedman Chapter 3University Key Equations Physics, 13/e Young/Freedman Chapter 3 Key Equations ˆ ˆ rjkxyz ˆ (position vector) (3.1) Instantaneous Velocity ˆ ˆ q The instantaneous rr velocity rjk rxyzis theˆ (position vector) (3.1) 21 (average velocity vector) (3.2) instantaneous av rate of change of tt21 t position vector with respectrr to r 21 (average velocity vector) (3.2) time. av tt t rrd 21 lim (instantaneous velocity vector) (3.3) t 0 tdt rrd lim (instantaneous velocity vector) (3.3) q The components oft 0thetdt dx dy dz instantaneous velocity are(components of instantaneous velocity) (3.4) xyzdt dt dt dx dy dz (components of instantaneous velocity) (3.4) xyzdt dt dt 21 q The instantaneous aav velocity(average of a acceleration vector) (3.8) tt21 t particle is always tangent21 to its path. aav (average acceleration vector) (3.8) tt21 t d a lim (instantaneous acceleration vector) (3.9) t 0 tdtd a lim (instantaneous acceleration vector) (3.9) t 0 tdt d d d aaax y z (components of instantaneous acceleration) (3.10) xyzd d d dt aaax dty dt z (components of instantaneous acceleration) (3.10) xyzdt dt dt

xt (cos)00 (projectile motion) (3.20) xt (cos)00 (projectile motion) (3.20)

ytgt(sin) 1 2 (projectile motion) (3.21) 002 1 2 ytgt(sin)002 (projectile motion) (3.21)

x 00cos (projectile motion) (3.22) x 00cos (projectile motion) (3.22)

y 00sin gt (projectile motion) (3.23) y 00sin gt (projectile motion) (3.23)

2 2 a rad a (uniform (uniform circular circular motion) motion) (3.28)(3.28) R rad R

ˆ ˆ rjkxyz ˆ (position vector) (3.1)

rr21 r av (average velocity vector) (3.2) tt21 t

rrd lim (instantaneous velocity vector) (3.3) Average tAcceleration0 tdt dx dy dz q The average accelerationduring a time(components interval of instantaneous∆" is velocity) (3.4) xyzdt dt dt defined as the velocity change ∆# divided by ∆".

21 aav (average acceleration vector) (3.8) tt21 t

d a lim (instantaneous acceleration vector) (3.9) t 0 tdt

d d d aaax y z (components of instantaneous acceleration) (3.10) xyzdt dt dt

xt (cos)00 (projectile motion) (3.20)

1 2 ytgt(sin)002 (projectile motion) (3.21)

x 00cos (projectile motion) (3.22)

y 00sin gt (projectile motion) (3.23)

2 a (uniform circular motion) (3.28) rad R

University Physics, 13/e ˆ ˆ rjkxyz ˆ (position vector) (3.1) Young/Freedman Chapter 3 Key Equations rr21 r av (average velocity vector) (3.2) tt21 t ˆ ˆ rjkxyz ˆ (position vector) (3.1) rrd lim (instantaneous velocity vector) (3.3) t 0 tdt rr21 r av (average velocity vector) (3.2) tt21 t dx dy dz xyz(components of instantaneous velocity) (3.4) dt dt dtrrd Instantaneous lim Acceleration (instantaneous velocity vector) (3.3) t 0 tdt q The instantaneous a 21 acceleration(averageis the acceleration vector) (3.8) av instantaneousdx ratett21 of dy change t dzof the (components of instantaneous velocity) (3.4) velocity withxyz dtrespect to dt time. dt d a lim (instantaneous acceleration vector) (3.9) t 0 tdt21 aav (average acceleration vector) (3.8) q Any particle following att21 curved tpath is accelerating,dx d y even dif zit has aaaxyz(components of instantaneous acceleration) (3.10) constantdt speed. dt dtd a lim (instantaneous acceleration vector) (3.9) q The components oft the0 tdt

instantaneous accelerationxt (cos)00 are (projectile motion) (3.20) dx d y dz aaaxyz(components of instantaneous acceleration) (3.10) dt dt1 dt2 ytgt(sin)002 (projectile motion) (3.21)

xt (cos)00 (projectile motion) (3.20) x 00cos (projectile motion) (3.22)

1 2 ytgt(sin)002 (projectile motion) (3.21) y 00sin gt (projectile motion) (3.23)

cos (projectile motion) (3.22) 2 x 00 a (uniform circular motion) (3.28) rad R y 00sin gt (projectile motion) (3.23)

Copyright © 2012 Pearson Education, Inc. Page 1 of 2 Acceleration Direction q Another useful way to think about instantaneous acceleration is in terms of its component parallel or perpendicular to the velocity. q Parallel component tells us about changes in the particles speed. q Perpendicular component tells us about changes in the particles direction of motion. Calculating Displacement

q A rover is exploring the surface of Mars, which we represent as a point. It has x- and y-coordinates that vary with time: x = 2.0 m - (0.25 m/s2 )t 2 y = (1.0 m/s)t + (0.025 m/s3 )t 3 q Find the rover’s coordinates and distance from the origin at t = 2 s. x = 2.0 m - (0.25 m/s2 )(2.0 s)2 =1.0 m y = (1.0 m/s)(2.0 s) + (0.025 m/s3 )(2.0 s)3 = 2.2 m r = x2 + y2 = 1.02 + 2.22 = 2.4 m Calculating Average Velocity q A rover is exploring the surface of Mars, which we represent as a point. It has x- and y-coordinates that vary with time: x = 2.0 m - (0.25 m/s2 )t 2 y = (1.0 m/s)t + (0.025 m/s3 )t 3 q Find the rover’s displacement and average velocity vectors for the interval t = 0.0 s to t = 2.0 s. !" = (2.0 m)+̂ + (0.0 m)-̂

!. = (1.0 m)+̂ + (2.2 m)-̂

∆! = !. − !" = (−1.0 m)+̂ + (2.2 m)-̂

∆! (−1.0 m)+̂ + (2.2 m)-̂ 2 = = =? 34 ∆5 2 s − 0 s Calculating Instantaneous Velocity q A rover is exploring the surface of Mars, which we represent as a point. It has x- and y-coordinates that vary with time: x = 2.0 m - (0.25 m/s2 )t 2 y = (1.0 m/s)t + (0.025 m/s3 )t 3 q Find a general expression for the rover’s instantaneous velocity vector V. dx v = = - (0.25 m/s2 )(2t) x dt dy v = =1.0 m/s + (0.025 m/s3 )(3t 2 ) y dt ! ˆ ˆ v = vxi + vy j q What is the rover’s instantaneous velocity at t = 2 s (magnitude and direction)? Calculating Acceleration q A rover is exploring the surface of Mars, which we represent as a point. It has x- and y-coordinates that vary with time: x = 2.0 m - (0.25 m/s2 )t 2 y = (1.0 m/s)t + (0.025 m/s3 )t 3 q Find a general expression for the rover’s instantaneous acceleration a. 2 vx = - (0.50 m/s )t dy v = =1.0 m/s + (0.075 m/s3 )(t 2 ) y dt dv a = x = -0.50 m/s2 x dt dv a = y = (0.075 m/s2 )(2t) y dt ! ˆ ˆ a = axi + ay j Summary I: Vectors q Vector addition

! + " = ('(*̂ + '+,)̂ + (.(*̂ + .+,)̂

=('(+.()*̂ + ('++.+),̂ q Vector subtraction " ! − " ! − " = ('(*̂ − '+,)̂ + (.(*̂ − .+,)̂ ! =('(−.()*̂ + ('+−.+),̂ University Physics, 13/e Young/FreedmanUniversity Physics, 13/e ChapterYoung/Freedman 3 Key Equations Chapter 3 Key Equations University Physics, 13/e ˆ ˆ rjkxyz ˆ (position vector) (3.1) University Physics, 13/e ˆ ˆ Young/Freedman rjkxyz ˆ (position vector) (3.1) University Physics, 13/e Young/Freedman rr r ChapterUniversity 3 Key Equations Physics, 13/e 21 (average velocity vector) (3.2) Chapter 3 Key Equationsav ttrr t r Young/Freedman Young/Freedman 2121 (average velocity vector) (3.2) Summaryav tt t II: Chapter 3 Key Equations ˆ 21ˆ Chapter 3 Key Equations rjkxyz ˆ (position vector) (3.1) rrd ˆ ˆ Position, limVelocity, (instantaneous rjkxyz Accelerationˆ velocity(position vector) vector) (3.3) (3.1) t 0 tdt rrˆ d ˆ rr lim rjkxyz rˆ (instantaneous(position vector) velocity vector) (3.1) (3.3) 21t 0 tdt q Position xyz ˆav ˆ ˆ (position(average vector) velocity vector) (3.1)(3.2) rjktt21 t rr21 r dx dy dzav (average velocity vector) (3.2) rr21 r(components of instantaneous velocity) (3.4) xyz tt21(average velocity t vector) (3.2) q Average velocitydtdx dtav dy dt dz tt21 t(components of instantaneous velocity) (3.4) xyzrrrr rd 21dtlim dt(average (instantaneous dt velocity velocity vector) vector) (3.2)(3.3) av t 0 tdt21 rrd q Instantaneous tt 21 velocity a t rr d lim(average acceleration (instantaneous vector) velocity vector) (3.8) (3.3) avlimt 0 (instantaneoustdt velocity vector) (3.3) t tt210 tdt t a 21 (average acceleration vector) (3.8) dx dyav dz tt21(components t of instantaneous velocity) (3.4) xyzrrd limdtdx dt dy (instantaneousdx dtd dz dy velocity dz vector) (3.3) t 0 xyz a limxyz (instantaneous(components (componentsaccelerationof instantaneous vector) of velocity) instantaneous velocity)(3.4)(3.9) (3.4) tdtdtt 0 dt dt dttdtd dt dt q Acceleration a lim (instantaneous acceleration vector) (3.9) 21t 0 aav tdt(average acceleration vector) (3.8) tt21 t d ad21d 21(average acceleration vector) (3.8) dx dy x dz av y a z(average acceleration vector) (3.8) aaaxyztt21av t(components of instantaneous acceleration) (3.4)(3.10) xyz(componentstt of instantaneous t velocity) dt dtdtd dt dtd dtd 21 ! ! x ! y z (components of instantaneous acceleration) (3.10) q r(t), aaa vxyz(t), and a(t)ared not necessarily along the same direction. dta lim dtd (instantaneous dt acceleration vector) (3.9) t a 0 lim tdt xt (cos) (instantaneous (projectile acceleration motion) vector) (3.9)(3.20) t 0 tdt00d 21 a lim (instantaneous acceleration vector) (3.9) t 0 a av (averagext (cos) accelerationtdt (projectile vector) motion) (3.8) (3.20) 00 tt21 t 1 2 dx d y d dz d x ytgty (sin)00dz 2 (projectile motion) (3.21) aaa xyz aaa(components(components of ofinstantaneous instantaneous acceleratio acceleration)n) (3.10)(3.10) dtxyz dt dt dt dtd dt 1 2 d xytgt(sin)y d z (projectile motion) (3.21) aaaxyz002 (components of instantaneous acceleration) (3.10) d dt cos dt (projectile dt motion) (3.22) a lim (instantaneousx 00 acceleration vector) (3.9) xt (cos) xt (cos)00 (projectile (projectile motion) motion) (3.20)(3.20) t 0 tdt 00 x 00cos (projectile motion) (3.22) xt (cos)00 (projectile motion) (3.20) y 00sin gt1 2 (projectile motion) (3.23) ytgt(sin) 2 (projectile motion) (3.21) ytgt(sin)001 2 (projectile motion) (3.21) d 002 dx y d z y 00sin gt (projectile motion) (3.23) 2 aaaxyz(components2 of instantaneous1 acceleration) (3.10) ytgt(sin)002 (projectile motion) (3.21) dt dt dt x 00cos (projectile motion) (3.22) arad cos (uniform (projectile circular motion) motion) (3.22)(3.28) x 00R 2 a rad sin (uniformgt (projectilecos circular motion) (projectile motion) motion) (3.23) (3.28) (3.22) y 00R x 00 xt (cos)00 (projectile motion) (3.20) y 00sin gt (projectile motion) (3.23) 2 sin (projectile motion) (3.23) a (uniformy 00 circulargt motion) (3.28) Copyright ytgt© 201(sin)2 Pearson Education,1rad2 2R Inc. (projectile motion) Page 1 of 2 (3.21) 00 a 2 (uniform circular motion) (3.28) rad R Copyright © 2012 Pearson Education, Inc. 2 Page 1 of 2 arad (uniform circular motion) (3.28) x 00cos (projectile motion)R (3.22) Copyright © 2012 Pearson Education, Inc. Page 1 of 2

Copyright © 2012 Pearson Education, Inc. Page 1 of 2 Calculating displacement q An ant moves in one direction with a displacement derived by A=3i+5j, then it turns to another direction describes as B=4i+10j. The units are in m. Find the angle in degrees of the final displacement of the ant relative to the X axis. (A) 79 (B) 11 (C) 65 (D) 25 (E) 36 Flying a plane q A plane is trying to land in the Newark International Airport from south at a reduced speed of 300 mph. However there is a strong wind blowing from west at 50 mph. In order to keep the plane in the north direction for landing, how many degrees away from due north (+ for east and – for west) should the captain steers the plane?

(A) -11.3 (B) -9.6 (C) 0 (D) 9.6 (E) 11.3 2-D Motion under Constant Acceleration q Motions in two dimensions are independent components q Constant acceleration equations in vector form of 1 ! = ! + %& ' − ' = ! & + %&+ !, = !+ + 2%(' − ' ) # # # 2 # # q Constant acceleration equations hold in each dimension

./ = .#/ + 0/& .3 = .#3 + 03& 1 1 1 − 1 = . & + 0 &+ 4 − 4 = . & + 0 &+ # #/ 2 / # #3 2 3 + + + + ./ = .#/ + 20/(1 − 1#) .3 = .#3 + 203(4 − 4#)

n t = 0 beginning of the process ! n a = a iˆ + a ˆj where a and a are constant x y ! x y ! n ˆ ˆ ˆ ˆ Initial velocity v0 = v0xi + v0 y j initial displacement r0 = x0i + y0 j Hints for solving problems q Define coordinate system. Make sketch showing axes, origin. q List known quantities. Find v0x, v0y, ax, ay, etc. Show initial conditions on sketch. q List equations of motion to see which ones to use. q Time t is the same for x and y directions.

x0 = x(t = 0), y0 = y(t = 0), v0x = vx(t = 0), v0y = vy(t = 0). q Have an axis point along the direction of ! if it is constant.

"# = "%# + '#( "0 = "%0 + '0( 1 1 ) − ) = " ( + ' (- 1 − 1 = " ( + ' (- % %# 2 # % %0 2 0 - - - - "# = "%# + 2'#() − )%) "0 = "%0 + 2'0(1 − 1%) Projectile Motion q A projectile is any body given an initial velocity that then follows a path determined by the effects of gravity. Projectile Motion: A Simple Case

q 2-D problem and define a coordinate system: x- horizontal, y- vertical (up +)

q Try to pick x0 = 0, y0 = 0 at t = 0 q Horizontal motion + Vertical motion

q Horizontal: ax = 0 , v0x = v0 2 q Vertical: ay = -g = -9.8 m/s , v0y = 0 q Equations: Horizontal Vertical

!" = !$" + &"' !/ = !$/ + &/' 1 1 ( − ( = ! ' + & ', 0 − 0 = ! ' + & ', $ $" 2 " $ $/ 2 /

, , , , !" = !$" + 2&"(( − ($) !/ = !$/ + 2&/(0 − 0$) Which ball strikes first? q Consider the motions of two golf balls shown in the right figure, the red one simply released and the yellow one shot horizontally. They have the same initial height and starting time. Which one will strike the ground first? (A) the red one (B) the yellow one (C) at same time (D) not enough information is given Projectile Motion: Example 1 q A ball rolls off a table of height h. The

ball has horizontal velocity v0 when it leaves the table. q How far away does it strike the ground? q How long does it take to reach the ground?

x0=?, y0=?, v0x=?, v0y=?, ax=?, ay=?

x=?, y=?, vx=?, vy=? q Vertical Displacement: x0 = 0, y0 = 0, v0x = v0, v0y = 0, 1 , − , = − '(. = −ℎ $ 2 ax = 0, ay = -g, y = -h q Time needed: ( = 2ℎ/' q Horizontal Displacement: !" = !$ !% = −'( / ( − /$ = !$( = !$ 2ℎ/' Projectile Motion: Generalized

q 2-D problem and define a coordinate system.

q Horizontal: ax = 0 and vertical: ay = -g.

q Try to pick x0 = 0, y0 = 0 at t = 0. q Velocity initial conditions:

n v0 can have x, y components. !"# = !" cos (" !") = !" sin ("

n vx is usually constant.

n vy changes continuously. q Equations: Horizontal Vertical

!# = !"# = !" cos (" !, = !") − ./ = !" sin (" − ./ 5 0 = 0 + ! / y= 4 + ! / − ./6 " "# " ", 6 = (!"cos (")/ 1 = (! sin ( )/ − ./6 " " 2 Projectile Motion q Provided air resistance is negligible, the horizontal component of the velocity remains constant q Vertical component of the acceleration is equal to the free fall acceleration –g q Vertical component of the velocity and the displacement in the y- direction are identical to those of a freely falling body q Projectile motion can be described !" = !$ cos ($ !) = !$ sin ($ − -. as a superposition of two 4 5 / = /$ + !$". y= 3 + ! . − -. independent motion in the x- and $ $) 5 = (!$cos ($). 1 y-directions = (! sin ( ). − -.5 $ $ 2 Mission Impossible 6 q Tom Cruise jumping off rooftop in Mission Impossible 6 (warning: spoil!) q https://www.youtube.com/watch?v=g_4y YnPEP70 Tom Cruise jumping off rooftop q In Mission Impossible 6, Tom Cruise jumps off a rooftop and landed on a nearby building that is 5 m lower. The horizontal distance between the two buildings is 10 m. Assume he jumps off the edge nearly horizontally. How fast in m/s should he run in order to make it to the other side (without dying)?

A) 5 m/s B) 10 m/s C) 20 m/s D) 7 m/s E) 15 m/s Summary III: Projectile Motion q Projectile motion is one type of 2-D motion under constant

acceleration, where ax = 0, ay = -g. q The key to analyzing projectile motion is to treat the x- and y- components separately and apply 1-D constant acceleration kinematics equations to each direction: horizontal direction Vertical direction

!" = !$" + &"' !/ = !$/ + &/' 1 1 ( − ( = ! ' + & ', 0 − 0 = ! ' + & ', $ $" 2 " $ $/ 2 / , , , , !" = !$" + 2&"(( − ($) !/ = !$/ + 2&/(0 − 0$)

&" = 0 &/ = −2 (projectile motion, no air resistance)