ch04-p088 - 88. Relative to the sled, the launch velocity...

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ground is v o = ( v ox v s ) i ^ + v oy j ^ . The horizontal and vertical displacement (relative to the ground) are therefore x land x launch = Δ x bg = ( v ox v s ) t flight y land y launch = 0 = v oy t flight + 1 2 ( g )( t flight ) 2 . Combining these equations leads to Δ x bg = 2 v ox v oy g © ¨ § ¹ ¸ · 2v oy g v s. The first term corresponds to the “ y intercept” on the graph, and the second term (in parentheses) corresponds to the magnitude of the “slope.” From Figure 4-54, we have 40 4 . bg s x v Δ=− This implies v oy = (4.0 s)(9.8 m/s 2 )/2 = 19.6 m/s, and that furnishes enough information to determine v ox . (a) v ox = 40 g /2 v oy = (40 m)(9.8 m/s 2 )/(39.2 m/s) = 10 m/s. (b) As noted above, v oy = 19.6 m/s.
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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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