Wednesday, 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!!1stterm 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!!!! Î Superposition of two motions– Horizontal motion with constant velocity and– Vertical motion under constant accelerationa =Gt=Show that a projectile motion is a parabola!!!xiv =fx=fy=2cos21cossin⎟⎟⎠⎞⎜⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛=iifiifiifvxgvxvyθθθPlug 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⎟⎟⎠⎞⎜⎜⎝⎛−=θθcosivιθyiv=siniivθxyai aj+=GGgj=−Gxivt= cosiivtθcosfiixvθ=21sin2iivtgtθ=−()21 2yivtgt+−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 xcomponent in 2-dim, because the ball is at y=0 position when it completed it’s flight.sec880or 0 ≈==∴gtt()80 0tgt−=()21402tgt+−=0m8sect∴≈xivt=()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 ARvt==vviiθhgvtiAιθsin=∴⎟⎟⎠⎞⎜⎜⎝⎛=gvRiiθ2sin22sin21sinsin⎟⎟⎠⎞⎜⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛=gvggvvyiiiiiifθθθ⎟⎟⎠⎞⎜⎜⎝⎛=gvyiif2sin22θWhat happens at the maximum height?sin2cosiiiivvgθθ⎛⎞⎜⎟⎝⎠yi yvat+=siniAvgtιθ−=0h =()212yivt gt=+−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?⎟⎟⎠⎞⎜⎜⎝⎛=gvhii2sin22θThis formula tells us that the maximum height can be achieved when θi=90o!!!⎟⎟⎠⎞⎜⎜⎝⎛=gvRiiθ2sin2This formula tells us that the maximum range can be achieved when 2θi=90o, i.e., θi=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 30oto 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 =cosivιθ20.0 cos30 17.3 /ms=× =Dsiniivθ20.0 sin30 10.0 /ms=× =D45.0−=212yivtgt−29.80 20.0 90.0tt−−=0xiv=cosivιθ=20.0 cos30 17.3 /ms×=Dyivgt−=siniivgtθ−=10.0 9.80 4.22 31.4 /ms−×=−22xf yfvv+=()2217.3 31.4 35.9 /ms=+−=Only physical
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