90 Motion in Two DimensionsP4.19(a)We use the trajectory equation:yxgxvffifii=−tancosθ2222.Withxf=36 0. m,vi=20 0. m s, and =°53 0.we findyf−°=36 053 09803602200530394222.tan....cos..mmsmm2afejbg.The ball clears the bar by394 3050889.−=m m.(b)The time the ball takes to reach the maximum height istvg120 053 0163==°=sin..ms sin±mss2.The time to travel 36.0 m horizontally is txvfix2=t236 020 053 0299=°=.(.cos..mms)saf.Since tt21>the ball clears the goal on its way down .P4.20The horizontal component of displacement is xvtvtfxiiicos. Therefore, the time required to
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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 .