89_Physics ProblemsTechnical Physics

# 89_Physics ProblemsTechnical Physics - 90 P4.19 Motion in...

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90 Motion in Two Dimensions P4.19 (a) We use the trajectory equation: yx gx v ff i f ii =− tan cos θ 2 22 2 . With x f = 36 0 . m , v i = 20 0 . m s, and 53 0 . we find y f ° = 36 0 53 0 980 360 2200 530 394 2 2 2 .t a n . .. .c o s . . m ms m m 2 af ej bg . The ball clears the bar by 394 305 0889 . −= m m . (b) The time the ball takes to reach the maximum height is t v g 1 20 0 53 0 163 == ° = sin . . ms s in ±ms s 2 . The time to travel 36.0 m horizontally is t x v f ix 2 = t 2 36 0 20 0 53 0 299 = ° = . (. c o s. . m ms) s a f . Since tt 21 > the ball clears the goal on its way down . P4.20 The horizontal component of displacement is xv t v t fx i i i cos . 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 .

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