p02_044

Fundamentals of Physics,Vol 1 (Chapters 1 - 20)

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44. There is no air resistance, which makes it quite accurate to set a = g = 9 . 8m/s 2 (where downward is the y direction) for the duration of the fall. We are allowed to use Table 2-1 (with ∆ y replacing x ) because this is constant acceleration motion; in fact, when the acceleration changes (during the process of catching the ball) we will again assume constant acceleration conditions; in this case, we have a 2 =+25 g = 245 m/s 2 . (a) The time of fall is given by Eq. 2-15 with v 0 =0and y =0. Thus, t = r 2 y 0 g = r 2(145) 9 . 8 =5 . 44 s . (b) The Fnal velocity for its free-fall (which becomes the initial velocity during the catching process) is found from Eq. 2-16 (other equations can be used but they would use the result from part (a)).
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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. 2-16, 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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