ch04-p107 - fence it will strike the ground in 0.5 s(so...

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107. (a) Eq. 2-15 can be applied to the vertical ( y axis) motion related to reaching the maximum height (when t = 3.0 s and v y = 0): y max y 0 = v y t 1 2 gt 2 . With ground level chosen so y 0 = 0, this equation gives the result y max = 1 2 g (3.0 s) 2 = 44 m. (b) After the moment it reached maximum height, it is falling; at t = 2.5 s, it will have fallen an amount given by Eq. 2-18: y fence y max = (0)(2.5 s) – 1 2 g (2.5 s) 2 which leads to y fence = 13 m. (c) Either the range formula, Eq. 4-26, can be used or one can note that after passing the
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Unformatted text preview: fence, it will strike the ground in 0.5 s (so that the total "fall-time" equals the "rise-time"). Since the horizontal component of velocity in a projectile-motion problem is constant (neglecting air friction), we find the original x-component from 97.5 m = v x (5.5 s) and then apply it to that final 0.5 s. Thus, we find v x = 17.7 m/s and that after the fence Δ x = (17.7 m/s)(0.5 s) = 8.9 m....
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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