113_Physics ProblemsTechnical Physics

113_Physics - 114*P4.72 Motion in Two Dimensions We follow the steps outlined in Example 4.7 eliminating t = d cos to find vi cos vi sin d cos gd 2

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114 Motion in Two Dimensions *P4.72 We follow the steps outlined in Example 4.7, eliminating t d v i = cos cos φ θ to find vd v gd v d i i i sin cos cos cos cos sin θφ −= 22 2 . Clearing of fractions, 2 2 vg d v ii cos sin cos cos cos sin θθφ . To maximize d as a function of , we differentiate through with respect to and set dd d = 0: 2 2 2 2 vv g dd d v i cos cos cos sin sin cos cos cos sin sin φθ +− = af . We use the trigonometric identities from Appendix B4 cos cos sin 2 θθθ =− and sin sin cos θθ = to find cos cos sin sin = . Next, sin cos tan = and cot tan 2 1 2 = give cot tan tan 29 0 2 φφ == ° so 90 2 and 45 2 . ANSWERS TO EVEN PROBLEMS P4.2 (a) ri j =+− 18 0 4 00 4 90 2 . ± .. ± tt t ej ; P4.8 (a) rij =+ 500 150 2 . ± . ± m; vij 300 . ± . ± t ms; (b) vi j 18 0 4 00 9 80 . ± . ± . t a f ; (c) aj 980 . ± ms 2 ; (b) j 10 0 6 00 . ± . ± m; 781 . m s (d) 54 0 32 1 . ± . ± m m a f a f ij ; (e) 18 0 25 4 . ± . ± bg ; P4.10 7 23 10 1 68 10 33 ×× m, m (f) . ± 2 j P4.12 (a) d g h 2 horizontally; P4.4 (a) j + 0 . ±± ω ai j 05 0 0 2 ± . ± 2 ; (b) tan F H G I K J 1 2 h d below the horizontal (b) rj = 400 . ± m
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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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