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1 Homework #4 Wrtten homework due now Onlne homework due on Tue Mar 3 by 8 am Exam 1 Answer keys and scores wll be posted by end of the week Supplemental Instructon sessons next week Wednesday 8:45 10:00 pm DuBos 10 th floor Thursday 4:15 5:30 pm (usual day and tme) 35 These 3 projectles have dentcal values of v y and all ht at the same tme. Battleshp trajectores These 3 projectles have dentcal values of v y and all ht at the same tme. Physcs
2 Balls from the same heght ht the ground at the same tme. Equal tme ntervals Why? Double drop Same ntal v y (both = 0) Dfferent v x s rrelevant! same acceleraton a y (both = -g) Physcs Equal tme ntervals A Double drop how fast B What was v or ball B just before t ht? v yf = v y gt = 9.8 m/s s = 6.3 m/s What was v x for ball B just before t ht? Same as t always has been: 2.5 m/s Physcs
3 Quz con toss Toss a con straght up and let t fall. At the top of the trajectory, s the acceleraton postve or negatve? At the top of the trajectory, s the velocty postve or negatve? Physcs A hunter PRS Monkey ams shoot hs cannon drectly at a monkey n a tree. When the cannon goes off, the startled monkey drops from the tree to the ground. Where does the cannon ball go? PRS A. The ball passes above the monkey B. The ball passes below the monkey C. The ball hts the monkey D. Not enough nformaton E. I have no dea Physcs
4 Monkey soluton Monkey: = y tree +v y Cannon ball wthout gravty: = 0 + v y t 1 2 gt 2 = y tree 1 2 gt 2 t = y tree = y +v ( y ) t a yt 2 Dfference due to gravty: y = 1 2 gt2 wth t = x tree v x Cannon ball wth gravty: = 0 + v y t 1 2 gt 2 = y tree 1 2 gt 2 y tree x tree Physcs Projectle Moton example In the move Road Trp, some students are seekng to jump a car across a gap n a brdge. One student, who professes to know what he s talkng about ( Of course I m sure wth physcs, I m always sure. ), says that they can easly make the jump. He gves the followng data: The car weghs 2100 pounds, wth passengers and luggage. Rght before the gap, there s a ramp that wll launch the car at an angle of 30.The gap s 10 feet wde. He then suggests that they should drve the car at a speed of 50 mph n order to make the jump. a. If the car actually went arborne at a speed of 50 mph at an angle of 30 wth respect to the horzontal, how far would t travel before landng? b. Does the mass of the car make any dfference n your calculaton? Physcs
5 Projectle Moton (see clp) Physcs Rght before the gap, there s a ramp that wll launch the car at an angle of 30.The gap s 10 feet wde. He then suggests that they should drve the car at a speed of 50 mph n order to make the jump. v = 50 mph = 22.2 m/s, = 30 o L = 3.0 m ( v x ) = v cos = (22.2 m/s)cos(30 o ) =19.2 m/s ( v y ) = v sn = (22.2 m/s)sn(30 o ) =11.1 m/s We need to fnd t to compute x va x = (v x ) t solve vertcal moton to get t Physcs
6 Vertcal moton = y + ( v y ) t 1 2 g( t)2 Horzontal moton x f = x + ( v x ) t 0 = 0 + ( v y ) t 1 2 g( t)2 0 = v y 1 2 gt ( v y ) = 1 2 gt t t = 2 ( v y) 2(11.1 m/s) = = 2.27 s g 9.8 m/s 2 x f = 0 + (19.2 m/s)(2.27 s) = 43.6 m x = 43.6 m s much greater than L = 10 ft = 3.0 m Physcs Projectle Moton: Delverng a Package by Ar (Hmwk #4) A relef arplane s delverng a food package to a group of people stranded on a very small sland. The sland s too small for the plane to land on, and the only way to delver the package s by droppng t. The arplane fles horzontally wth constant speed of 250 mph at an alttude of 800 m. The postve x and y drectons are defned n the fgure. For all parts, assume that the "sland" refers to the pont at a dstance D from the pont at whch the package s released, as shown n the fgure. Ignore the heght of ths pont above sea level. Assume that the acceleraton due to gravty s g = 9.80 m/s 2. A. Tme for package to reach sea level? B. At what dstance D should package be released for t to land on the sland? C. Speed v f of the package when t hts the ground? Physcs
7 Projectle Moton: Delverng a Package by Ar Know: v 0 = 250 mph and h = 800 m Fnd: tme t, dstance D, speed v f Prepare the problem: Draw pcture (ntal + fnal snapshots) Organze known quanttes Solve usng knematcs equatons treat moton along x- and y-axes separately Frst solve vertcal moton to get t Then fnd horzontal dsplacement durng that t 2 2 Fnd (v y ) f to compute fnal speed v f = ( v x ) f + ( v y ) f Physcs Projectle Moton: Speed of a Softball (Hmwk #4) A softball s ht over a thrd baseman's head wth some speed v 0 at an angle theta above the horzontal. Immedately after the ball s ht, the thrd baseman turns around and begns to run at a constant velocty v = 7.00 m/s. He catches the ball t = 2.00 s later at the same heght at whch t left the bat. The thrd baseman was orgnally standng L = 18.0 m from the locaton at whch the ball was ht. A. Fnd v 0 (use g = 9.81 m/s 2 ). B. Fnd the angle n degrees. Use t = 2.0 s to fnd (v y ) Compute horzontal dsplacement to fnd (v x ) Physcs
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