Unformatted text preview: v = − q v 2 − 2 g ( y − y ) = − p 2 gy = − 53 . 3 m / s where the negative root is chosen since this is a downward velocity. (c) ±or the catching process, the answer to part (b) plays the role of an initial velocity ( v = − 53 . 3 m/s) and the Fnal velocity must become zero. Using Eq. 216, we Fnd ∆ y 2 = v 2 − v 2 2 a 2 = − ( − 53 . 3) 2 2(245) = − 5 . 80 m where the negative value of ∆ y 2 signiFes that the distance traveled while arresting its motion is downward....
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 Acceleration, Velocity, initial velocity, final velocity, constant acceleration motion, constant acceleration conditions

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