114. We neglect air resistance, which justifies setting a= –g= –9.8 m/s2(taking downas the –ydirection) for the duration of the motion of the shot ball. We are allowed to use Table 2-1 (with Δyreplacing Δx) because the ball has constant acceleration motion. We use primed variables (except t) with the constant-velocity elevator (so '10m/sv=), and unprimed variables with the ball (with initial velocity 02030 m/svv′=+ =, relative to the ground). SI units are used throughout. (a) Taking the time to be zero at the instant the ball is shot, we compute its maximum heighty(relative to the ground) with 22002()gyy=−−, where the highest point is characterized by v= 0. Thus, yyvg=+=om02276whereoo230m′=(where o28 my′ =is given in the problem) and v0= 30 m/s relative to the ground as noted above. (b) There are a variety of approaches to this question. One is to continue working in the frame of reference adopted in part (a) (which treats the ground as motionless and “fixes”
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.