Physics 111. Lecture 7 (Walker: 4.2-5) 2D Motion Examples Projectile Motion

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1 Physics 111 Lecture 7 (Walker: 4.-5) D Motion Eamples Projectile Motion Sept. 16, 9 -D Motion -- Constant Acceleration r r r r = v t at t v t a t y y yt y v t at r r r v = v at v = v a t v = v a t y y y v = v a v = v a y y y y Lecture 7 1/7 Lecture 7 /7 Eample: Hummer Accelerates Humminbird is initially movin vertically with speed of 4.6 m/s and acceleratin horizontally at a constant rate of 11 m/s. Find the horizontal and vertical distance throuh which it moves in.55 s. 1 = v t at = at = (11 m/s )(.55 s) = 1.7 m y = y vyt a yt = vyt = (4.6 m/s)(.55 s) =.5 m Projectile Motion (Objects with vertical acceleration & constant horizontal velocity) Assumptions: Inore air resistance Use = 9.8 m/s, direction downward, for free fall Inore the Earth s rotation If the y-ais points upward, the acceleration in the - direction is zero and the acceleration in the y- direction is 9.8 m/s durin the projectile motion. Lecture 7 3/7 Lecture 7 4/7

2 Equations: Constant Acceleration with a = and a y = - t v t v = v Projectile Motion: Basic Equations The acceleration is independent of direction of velocity: yt y v yt t v = v t y y v = v y y y Lecture 7 5/7 Lecture 7 6/7 Independence of Vertical and Horizontal Motion When you drop a ball while runnin with constant velocity, it appears to you to drop straiht down from the point where you released it. To a person at rest, the ball follows a curved path that combines horizontal and vertical motions. Launch Anle Launch anle: direction of initial velocity with respect to horizontal Lecture 7 7/7 Lecture 7 8/7

3 Zero Launch Anle In the zero launch anle case, the initial velocity in the y-direction is zero. Here are the equations of motion, with = and y = h: Zero-Launch Trajectory This is the trajectory (path in space) of a projectile launched horizontally from y = 9.5m Lecture 7 9/7 Lecture 7 1/7 Eample: Cliff Divin Geore and Sam dive from a hih overhanin cliff into a lake. Geore drops straiht down. Sam runs horizontally and dives outward. (a) It they leave the cliff at the same time, which boy reaches the water first? Since vertical motion determines time to reach the water, both boys reach the water at the same. (b) Which boy hits the water with the reater speed? Geore reaches water with only vertical velocity; Sam reaches the water with both horizontal and vertical velocity components. Vertical velocities same for both, so Sam s speed at water reater than Geore s. Lecture 7 11/7 Eample: Electron in Motion Electron travelin with horizontal speed of cm/s oes between deflection plates that ive it an upward acceleration of cm/s. After that, (a) How lon does it take for electron to cover a horizontal distance of 6. cm? (b) What is the vertical displacement durin this time? t v t a t v t = (6. cm) t = = = 9 v (.1 1 cm/s) y y = y yt y v t at at s y = a t = (5.3 1 cm/s )(.95 1 s) =.31 cm y Lecture 7 1/7

4 Projectile - General Launch Anle In eneral, v = v cos θ and v y = v sin θ (This ASSUMES θ is measured CCW from ais!) The equations of motion are: General Launch Anle Trajectory Snapshots of a trajectory; red dots are at t = 1 s, t = s, and t = 3 s = v t y = y v y t - ½t Lecture 7 13/7 Lecture 7 14/7 Eample: A Home Run Baseball leaves bat makin 3 anle with round. It crosses fence 1 m from home plate at same heiht that it was struck. (Nelect air resistance.) What was its velocity as it left bat? Lecture 7 15/7 v = v cosθ =.866v 1 = 1m =.866v t 1 so t 1 = 1m/(.866v ) v y = v sinθ =.5v y 1 = = v y t 1 -(.5)t 1 = (.5v )[(1m/(.866v )]-(4.9m/s )[1m/(.866v )] Thus 57.7m = (6533m 3 /s )/v v = 33.6 m/s Lecture 7 16/7

5 Eample: Catch a Thief Police officer chases thief across rooftops. They are both runnin when they come to a ap between buildins that is 4. m wide and has a drop of 3. m. Thief, havin studied physics, leaps at 5. m/s at anle of 45 above horizontal and clears ap. Officer did not study physics and thinks he should maimize his horizontal velocity, so leaps horizontally at 5. m/s. (a) Does he clear the ap? (b) By how much does the thief clear the ap? (a) Does he clear the ap? No! (b) By how much does thief clear the ap? y = y vyt t Police: Thief: 3. m = (9.81 m/s )t t = 6. m / 9.81 m/s =.78 s = vt = (5. m/s)(.78 s) = 3.91 m 3. m = (5. m/s)sin 45 t (9.81 m/s ) t t =.5 s or t = 1. s = vt = (5. m/s)(1. s)cos45 = 4.31 m Lecture 7 17/7 Lecture 7 18/7 Projectile Motion: Rane Rane: the horizontal distance a projectile travels If the initial and final elevation are the same: Projectile Motion:Maimum Rane The rane is a maimum when θ = 45 : Lecture 7 19/7 Lecture 7 /7

6 Projectile Motion: Symmetry Symmetry in projectile motion: Path of Projectile Same y and v Same y and v y Lecture 7 1/7 Lecture 7 /7 ConcepTest A un is accurately aimed at criminal hanin from utter of buildin. The taret is well within un s rane, but the instant un is fired and bullet moves with a speed v o, criminal lets o and drops to round. What happens? The bullet hits criminal reardless of value of v o.. hits criminal only if v o is lare enouh. 3. misses criminal. Eample: A Supply Drop Helicopter drops supply packae to flood victims on raft. When packae is released, helicopter is 1 m directly above raft and flyin at velocity 5. m/s at anle 36.9 above horizontal. [Nelect air resistance]. (a) How lon is packae in air? (b) How far from raft does packae land? Lecture 7 3/7 Lecture 7 4/7

7 v = v sin θ = (5. m/s)(sin36.9 ) = 15. m/s; v = v cos θ =. m/s y y() t = v yt t = -1m = (15. m/s)t - (.5)(9.8 m/s )t t (15. m/s) ± (15. m/s) (9.81 m/s )( 1 m) (9.81 m/s ) t = 3.4 s and t = 6.3 s = h = v t = (. m/s)(6.3 s) = 16 m The packae will hit the water 16 m from raft. Lecture 7 5/7 Summary - D Kinematics Components of motion in the - and y- directions can be treated independently. In projectile motion, the y-component of acceleration is ; there is no - acceleration. If the launch anle is zero, the initial velocity has only an -component. The path followed by a projectile is a parabola. The rane is the horizontal distance the projectile travels. Lecture 7 6/7 End of Lecture 7 Before the net lecture, Walker Homework Assinment #4a should be submitted usin WebAssin by 11: PM on Saturday, Sept. 19. Lecture 7 7/7