PHYS 1441 – Section 002 Lecture #7AnnouncementsSlide 3Special Project for Extra CreditWhat is the Projectile Motion?Projectile MotionKinematic Equations for a projectile motionSlide 8Example for a Projectile MotionEx.3.9 The Height of a KickoffFirst, the initial velocity componentsMotion in y-direction is of the interest..Now the nitty, gritty calculations…Ex.3.9 extended: The Time of Flight of a KickoffWhat is y when it reached the max range?Now solve the kinematic equations in y direction!!Ex.3.9 The Range of a KickoffSlide 18Example cont’dHorizontal Range and Max HeightSlide 21Maximum Range and HeightWednesday, Sept. 29, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu1PHYS 1441 – Section 002Lecture #7Wednesday, Sept. 29, 2010Dr. Jaehoon Yu•Projectile Motion•Maximum Range and Height•Force•Newton’s Second LawAnnouncements•First term exam results–Class Average: 52/96•Equivalent to: 55/100–Top score: 94/96•Evaluation policy–Homework: 30%–Final comprehensive exam: 25%–Better of the two non-comprehensive term exam: 20%–Lab: 15%–Quizzes: 10%–Extra credit: 10% •Colloquium today at 4pm in SH101Wednesday, Sept. 29, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu2Wednesday, Sept. 29, 20103PHYS 1441-002, Fall 2010 Dr. Jaehoon YuWednesday, Sept. 29, 20104Special Project for Extra Credit•Show that the trajectory of a projectile motion is a parabola!!–20 points–Due: Wednesday, Oct. 6–You MUST show full details of your OWN computations to obtain any credit•Beyond what was covered in page 8 of this lecture note!!PHYS 1441-002, Fall 2010 Dr. Jaehoon YuWednesday, Sept. 29, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu5What is the Projectile Motion?•A 2-dim motion of an object under the gravitational acceleration with the following assumptions–Free fall acceleration, g, is constant over the range of the motion• • and–Air resistance and other effects are negligible•A motion under constant acceleration!!!! Superposition of two motions–Horizontal motion with constant velocity ( no acceleration )–Vertical motion under constant acceleration ( g ) rg = −9.8rj m s2( ) ax= ay= 0 m s2 −9.8m s2 vxf= vyf= vx0 vy0+ayt = vy0+ −9.8( )tWednesday, Sept. 29, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu6Projectile Motion The only acceleration in this motion. It is a constant!!Maximum heightWednesday, Sept. 29, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu7x-component Δx=vxot Δx=vxot vx=vxo vx02=vxo2 Δy=12vyo+vy( )t Δy=vyot −12gt2 vy=vyo−gt vy2=vyo2−2gyKinematic Equations for a projectile motiony-component ax= ay= 0 −gur= −9.8m s2Wednesday, Sept. 29, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu8 ra =t =xfvicosθiShow that a projectile motion is a parabola!!! vxi=xf= yf=vyit +12−g( ) t2 yf=Plug t into the aboveIn a projectile motion, the only acceleration is gravitational one whose direction is always toward the center of the earth (downward).ax=0What kind of parabola is this?yf=xftanθi−g2vi2cos2θi⎛⎝⎜⎞⎠⎟xf2 vyi=x-component y-component axri +ayrj = −grjvxit =vicosθit visinθixfvicosθi⎛⎝⎜⎞⎠⎟ −12gxfvicosθi⎛⎝⎜⎞⎠⎟2 =visinθit −12gt2 vicosθι visinθiWednesday, Sept. 29, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu9Example for a Projectile MotionA ball is thrown with an initial velocity vv=(20ii+40jj)m/s. Estimate the time of flight and the distance from the original position when the ball lands.Which component determines the flight time and the distance? yf= xf=Flight time is determined by the y component, because the ball stops moving when it is on the ground after the flight.Distance is determined by the x component in 2-dim, because the ball is at y=0 position when it completed it’s flight.∴ t = 0 or t =80g≈ 8 sec t 80 −gt( )=0 40t +12−g( )t2= 0m ∴ t ≈8sec vxit = 20 ×8=160 m( )So the possible solutions are…Why isn’t 0 the solution?Wednesday, Sept. 29, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu10A placekicker kicks a football at an angle of 40.0 degrees andthe initial speed of the ball is 22 m/s. Ignoring air resistance, determine the maximum height that the ball attains.Ex.3.9 The Height of a KickoffWednesday, Sept. 29, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu11 v0 y= v0 x=First, the initial velocity components v0=22m s v0 y v0 x θ =40oov cosθ = 22 m s( )cos40o= 17 m sov sinθ = 22 m s( )sin40o= 14 m sWednesday, Sept. 29, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu12yayvyv0ytMotion in y-direction is of the interest..-9.8 m/s2+14 m/s0 m/s?Wednesday, Sept. 29, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu13yayvyv0yt? -9.80 m/s20 14 m/s2yv =Solve for y y =vy2−voy22ay y =0− 14m s( )22 −9.8m s2( )=+10 mNow the nitty, gritty calculations…Which kinematic formula would you like to use? What happens at the maximum height?The ball’s velocity in y-direction becomes 0!!And the ball’s velocity in x-direction? Stays the same!! Why?Because there is no acceleration in x-direction!! voy2+2ayyWednesday, Sept. 29, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu14What is the time of flight between kickoff and landing?Ex.3.9 extended: The Time of Flight of a KickoffWednesday, Sept. 29, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu15yayvyvoyt?What is y when it reached the max range?-9.80 m/s214 m/s0 mWednesday, Sept. 29, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu16yayvyvoyt0 -9.80 m/s214 m/s ?Now solve the kinematic equations in y direction!! y =voyt +12ayt2 0 =voyt +12ayt2=0t = voy+12ayt =0t =Since y=0orSolve for t −2voyay= −2⋅14 −9.8=2.9s −voy12ay= t voy+12ayt( )Two soultionsWednesday, Sept. 29, 2010PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu17Calculate the range R of the projectile.x =Ex.3.9 The Range of a Kickoff voxt +12axt2= voxt = 17 m s( )2.9 s( )=+49 mWednesday, Sept. 29, 201018PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu18Example for a Projectile Motion•A stone was thrown upward from the top of a cliff at an angle of 37o to horizontal with initial speed of 65.0m/s. If the height of the cliff is 125.0m, how long is it before the stone hits the ground? xivfy = gt2−78.2t −250=t =78.2± −78.2( )2−4×9.80×(−250)2×9.80t =−2.43s or t =10.4st =10.4syivSince negative time does not exist. =vicosθι =65.0×cos37o=51.9m/ s =visinθi =65.0×sin37o=39.1m/ s −125.0= vyit −12gt2 9.80t2−78.2t −250=0BecomesWednesday, Sept. 29, 201019PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu19Example
View Full Document
Unlocking...