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**Unformatted text preview: **mg where . 2 . 2 = For constant speed, . 2.2 cos 2.2 sin 2 r =-- Dv mg mg Solving for v gives . s m 29 = v c) For angle cos sin , 2 r =--Dv mg mg and D mg v ) cos (sin r-= The terminal speed for a falling object is derived from , 2 t =-mg Dv so . t D mg v = v v cos sin r t-= And since v v cos ) 015 . ( sin , 015 . t r-= =...

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