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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 y y v t a t f i yi y = + + 1 1 2 1 2 0 0 25 0 70 0 1 2 9 80 2 25 0 70 0 9 80 4 79 1 1 2 1 = + ° = ° = . sin . . ( . )sin . . . . m s m s m s m s s 2 2 b gb g e j t t t The distance to your target is x x v t f i xi = = ° = 1 25 0 70 0 4 79 41 0 . cos . . . m s s m b g a f . Now the second snowball we describe by y y v t a t f i yi y = + + 2 2 2 1 2 0 25 0 4 90 2 2 2 2 = . sin . m s m s 2 b g e j θ t t t 2 2 5 10 = . sin s a f θ x x v t f i xi = 2 41 0 25 0 5 10 128 2 2 2 2 . . cos . sin sin cos m m s s m = = b g a f a f θ θ θ θ 0 321 2 2 . sin cos = θ θ Using sin sin cos 2 2 θ θ θ = we can solve 0 321 1 2 2 2 . sin = θ 2 0 643 2 1 θ = sin . and θ 2 = ° 20 0 . . (b) The second snowball is in the air for time
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