ch02-p045 - 45. We neglect air resistance, which justifies...

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2 2 (2 4m / s ) 29.4 m 2(9.8 m/s ) y Δ=− =− so that it fell through a height of 29.4 m. (b) Solving v = v 0 gt for time, we find: 0 2 0(2 4m / s ) 2.45 s. 9.8 m/s vv t g −− == =
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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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