ch04-p128 - 128. The figure offers many interesting points...

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point shown (1.25 s after the ball is released) which is when the ball returns to its original height. In English units, g = 32 ft/s 2 . (a) Using x x 0 = v x t we obtain v x = (40 ft)/(1.25 s) = 32 ft/s. And yy v t g t y −= = 00 1 2 2 0 yields () 2 1 0 2 32 ft/s 1.25 s 20 ft/s. y v == Thus, the initial speed is 22 | | (32 ft/s) (20 ft/s) 38 ft/s. vv + = G (b) Since v y = 0 at the maximum height and the horizontal velocity stays constant, then the speed at the top is the same as v x = 32 ft/s. (c) We can infer from the figure (or compute from
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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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