210020.fyfpfyyvtgt−==−This gives us v0y= 1.25gp, and we see we need another equation (by analyzing another point, say, the next-to-last one) 212ypytgt−withy= 6 and t= 2; this produces our second equation v0y= 2 + gp. Simultaneous solution of these two equations produces results for v0yand gp(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 gp= 8.0 m/s2.(c) Solving for tg(the time to reach the ground) in yyvtgtgygpg== +−0122leads to a positive answer: tg= 2.7 s. (d) With g= 9.8 m/s2, the method employed in part (c) would produce the quadratic equation24.91020ggtt−++=and then the positive result tg= 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.