ch04-p132 - 132. We let gp denote the magnitude of the...

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2 1 00 2 0. f yf pf y yv t g t −= = This gives us v 0 y = 1.25 g p , and we see we need another equation (by analyzing another point, say, the next-to-last one) 2 1 2 yp y t g t with y = 6 and t = 2; this produces our second equation v 0 y = 2 + g p . Simultaneous solution of these two equations produces results for v 0 y and g p (relevant to part (b)). Thus, our complete answer for the initial velocity is ˆˆ (10 m/s)i (10 m/s)j . v =+ G (b) As a by-product of the part (a) computations, we have g p = 8.0 m/s 2 . (c) Solving for t g (the time to reach the ground) in yy v t g t g y gp g == + 0 1 2 2 leads to a positive answer: t g = 2.7 s. (d) With g = 9.8 m/s 2 , the method employed in part (c) would produce the quadratic equation 2 4.9 10 2 0 gg tt −+ + = and then the positive result t g = 2.2 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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