(b) Since 220(19 m/s)31 m/syv+=(the first point on the graph), we find 024.5 m/s.yv=Thus, with t= 2.5 s, we can use 21max002yyyvtgt−= −orvvgyyyy202002== −−max,bgor ()1max002yyyyvvt−=+to solve. Here we will use the latter: max00max11() (0 24.5m/s)(2.5 s)31 mvvty+¡=+=where we have taken y0= 0 as the ground level. 42. In this projectile motion problem, we have v0= vx= constant, and what is plotted is
This is the end of the preview. Sign up
access the rest of the document.