Chapter 223Q2.17Above. Your ball has zero initial speed and smaller average speed during the time of flight to thepassing point.SOLUTIONS TO PROBLEMSSection 2.1Position, Velocity, and SpeedP2.1(a)v=230. ms(b)vxt==m ms= 16.1 m s∆∆57 59 20300...−(c)vxt−=∆∆57 5011 5..m5.00 sms*P2.2(a)vxtFHGIKJ×FHGIKJ=×−∆∆20113156 1021077ft1 yrm3.281 ftyrs.or in particularly windy timesvxtFHGIKJ×FHGIKJ−∆∆1001111076ft1 yrm3.281 ftyrs..(b)The time required must have been∆∆txvFHGIKJFHGIKJ3 0001 6091051038mi10 mm yrm1 mimm1 myr .P2.3(a)vxt=∆∆105m2 s(b)v512m4 s
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .