90_Physics ProblemsTechnical Physics

90_Physics ProblemsTechnical Physics - Chapter 4 *P4.21 (a)...

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Chapter 4 91 *P4.21 (a) For the horizontal motion, we have xx v t a t v v fix i x i i =+ + =+ ° + = 1 2 24 0 53 2 2 0 18 1 2 m s ms cos . .. af a f (b) As it passes over the wall, the ball is above the street by yy v t a t fiy i y 1 2 2 y f ° + − = 01 8 1 5 32 2 1 2 98 22 813 2 .s i n . . . . s s m 2 bg a f ej . So it clears the parapet by 7 113 m m −= . (c) Note that the highest point of the ball’s trajectory is not directly above the wall. For the whole flight, we have from the trajectory equation yx g v x fi f ii f =− F H G I K J tan cos θ 2 2 or 65 3 2181 53 2 2 2 m 2 ° F H G G I K J J tan . .c o s ff . Solving,
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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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