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UT Arlington PHYS 1441 - Displacement, Velocity, and Acceleration in 2-dim

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PHYS 1441 – Section 004 Lecture #7Displacement, Velocity, and Acceleration in 2-dimProjectile MotionExample for Projectile MotionHorizontal Range and Max HeightMaximum Range and HeightSlide 7Wednesday, Feb. 11, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu1PHYS 1441 – Section 004Lecture #7Wednesday, Feb. 11, 2004Dr. Jaehoon Yu–Projectile Motion–Maximum Range and Height–Reference Frame and Relative Velocity•Chapter four: Newton’s Laws of Motion–Newton’s First Law of Motion–Newton’s Second Law of Motion–Gravitational Force–Newton’s Third Law of MotionToday’s homework is homework #4, due 1pm, next Wednesday!!1st term exam in the class at 1pm, Monday, Feb. 23!!Wednesday, Feb. 11, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu2Displacement, Velocity, and Acceleration in 2-dim•Displacement:ifrrr •Average Velocity:ififttrrtrv•Instantaneous Velocity:dtrdtrvt 0lim•Average Accelerationififttvvtva•Instantaneous Acceleration:220limdtrddtrddtddtvdtvatHow is each of these quantities defined in 1-D?Wednesday, Feb. 11, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu3Projectile Motion•A 2-dim motion of an object under the gravitational acceleration with the assumptions–Gravitational cceleration, -g, is constant over the range of the motion–Air resistance and other effects are negligible•A motion under constant acceleration!!!!  Superposition of two motions–Horizontal motion with constant velocity and–Vertical motion under constant accelerationa =rt =Show that a projectile motion is a parabola!!!xiv =fx =fy =2cos21cossiniifiifiifvxgvxvyPlug in the t 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?222cos2tanfiiiffxvgxycosiviqyiv =sini iv qx ya i a j+ =r rg j=-rxiv t =cosi iv tqcosfi ixv q=21sin2i iv t gtq= -( )21 2yiv t g t+ -Wednesday, Feb. 11, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu4Example for Projectile MotionA ball is thrown with an initial velocity vv=(20ii+40jj)m/s. Estimate the time of flight and the distance the ball is from the original position when landed.Which component determines the flight time and the distance?fy =fx =Flight time is determined by y component, because the ball stops moving when it is on the ground after the flight.Distance is determined by x component in 2-dim, because the ball is at y=0 position when it completed it’s flight.sec880or 0 gtt( )80 0t gt- =( )21402t g t+ - =0m 8sect\ �xiv t =( )20 8 160 m� =Wednesday, Feb. 11, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu5Horizontal Range and Max Height•Based on what we have learned in the previous pages, one can analyze a projectile motion in more detail–Maximum height an object can reach–Maximum rangeyfv =At the maximum height the object’s vertical motion stops to turn around!!fy =Since no acceleration in x, it still flies even if vy=02xi AR v t= =vviihgvtiAsingvRii2sin22sin21sinsingvggvvyiiiiiifgvyiif2sin22What happens at the maximum height?sin2 cosi ii ivvgqq� �� �� �yi yv a t+ = sini Av gtiq - =0h =( )212yiv t g t= + -Wednesday, Feb. 11, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu6Maximum Range and Height•What are the conditions that give maximum height and range of a projectile motion?gvhii2sin22This formula tells us that the maximum height can be achieved when i=90o!!!gvRii2sin2This formula tells us that the maximum range can be achieved when 2i=90o, i.e., i=45o!!!Wednesday, Feb. 11, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu7Example for Projectile Motion•A stone was thrown upward from the top of a building at an angle of 30o to horizontal with initial speed of 20.0m/s. If the height of the building is 45.0m, how long is it before the stone hits the ground? xiv =•What is the speed of the stone just before it hits the ground? yfv =fy =220.0 90.0gt t- - = 80.92)90(80.94200.202tstst 22.4or 18.2 st 22.4v =xfv =yiv =cosiviq20.0 cos30 17.3 /m s= � =osini iv q20.0 sin 30 10.0 /m s= � =o45.0- =212yiv t gt-29.80 20.0 90.0t t- - =0xiv =cosiviq =20.0 cos30 17.3 /m s� =oyiv gt- = sini iv gtq - =10.0 9.80 4.22 31.4 /m s- � =-2 2xf yfv v+ =( )2217.3 31.4 35.9 /m s= + - =Only physical


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UT Arlington PHYS 1441 - Displacement, Velocity, and Acceleration in 2-dim

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