ch02-p088 - it started (the top of the building). (c) Now...

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88. We adopt the convention frequently used in the text: that "up" is the positive y direction. (a) At the highest point in the trajectory v = 0. Thus, with t = 1.60 s, the equation v = v 0 gt yields v 0 = 15.7 m/s. (b) One equation that is not dependent on our result from part (a) is y y 0 = vt + 1 2 gt 2 ; this readily gives y max y 0 = 12.5 m for the highest ("max") point measured relative to where
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Unformatted text preview: it started (the top of the building). (c) Now we use our result from part (a) and plug into y y = v t + 1 2 gt 2 with t = 6.00 s and y = 0 (the ground level). Thus, we have 0 y = (15.68 m/s)(6.00 s) 1 2 (9.8 m/s 2 )(6.00 s) 2 . Therefore, y (the height of the building) is equal to 82.3 m....
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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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