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113_Physics ProblemsTechnical Physics

# 113_Physics ProblemsTechnical Physics - 114*P4.72 Motion in...

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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 v d v gd v d i i i sin cos cos cos cos sin θ φ θ φ θ φ = − 2 2 2 2 2 . Clearing of fractions, 2 2 2 2 2 2 v gd v i i cos sin cos cos cos sin θ θ φ φ θ φ = − . To maximize d as a function of θ , we differentiate through with respect to θ and set dd d θ = 0 : 2 2 2 2 2 2 2 2 v v g dd d v i i i cos cos cos sin sin cos cos cos sin sin θ θ φ θ θ φ θ φ θ θ φ + = − a f a f . We use the trigonometric identities from Appendix B4 cos cos sin 2 2 2 θ θ θ = and sin sin cos 2 2 θ θ θ = to find cos cos sin sin φ θ θ φ 2 2 = . Next, sin cos tan φ φ φ = and cot tan 2 1 2 θ θ = give cot tan tan 2 90 2 φ φ θ = = °− a f so φ θ = °− 90 2 and θ φ = °− 45 2 . ANSWERS TO EVEN PROBLEMS P4.2 (a) r i j = + 18 0 4 00 4 90 2 . ± . . ± t t t e j ; P4.8 (a) r i j = + 5 00 1 50 2 . ± . ± t t e j m; v i j = + 5 00 3 00 . ± . ± t e j m s ; (b) v i j = + 18 0 4 00 9 80 . ± . ± . t a f ; (c) a j = − 9 80 . ± m s 2 e j ; (b) r i j = + 10 0 6 00 . ± . ± e j m; 7 81 . m s (d) 54 0 32 1 . ± . ± m m a f a f i j ; (e) 18 0 25 4 . ± . ± m s m s b g b g i j ; P4.10 7 23 10 1 68 10 3 3 . . × × m, m e j (f) 9 80 . ± m s 2 e j j P4.12 (a) d g h 2 horizontally;
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