86_Physics ProblemsTechnical Physics

# 86_Physics ProblemsTechnical Physics - Chapter 4 P4.13 (a)...

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Chapter 4 87 P4.13 (a) The time of flight of the first snowball is the nonzero root of yy v t a t fiy i y =+ + 11 2 1 2 00 2 5 0 7 0 0 1 2 980 2250 700 479 2 1 ° − = ° = .s i n . . (. ) s i n. . .. ms ±ms s 2 2 bg b g ej tt t The distance to your target is xxv t fix i −= = ° = 1 25 0 70 0 4 79 41 0 .c o s . . . s m af . Now the second snowball we describe by v t a t i y + 22 2 1 2 02 5 0 4 9 0 2 2 =− i n . 2 θ t 510 = i n s t i 2 41 0 25 0 5 10 128 2 2 . . cos . sin sin cos m m s s m == bg af a f θθ 0321 i n c o s = Using sin sin cos = we can solve 1 2 2 2 i n = 20 6 4 3 2 1 = sin . and 2 20 0 (b) The second snowball is in the air for time t 5 10 5 10 20 1 75 ° = i n i n . s s , so you throw it after the first by
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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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