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Unformatted text preview: at the time of the explosion). m d 2 z dt 2 = − mg, z (0) = H, ˙ z (0) = v o The solution is z ( t ) = H + v o t − 1 2 gt 2 The center of mass is defined by z ( t ) = 1 m } 2 5 mz A ( t ) + 3 5 mz B ( t ) ] Now, Part A strikes the ground when z A ( τ ) = 0 , and this equation simplifies to z B ( τ ) = 5 3 z ( τ ) = 5 3 H + 5 3 v o τ − 5 6 g τ 2 We are given z B ( τ ) = 5 6 H . Thus, 5 6 H = 5 3 H + 5 3 v o τ − 5 6 g τ 2 = ⇒ 5 3 v o τ − 5 6 g τ 2 − 5 6 H = 0 Solving for v o yields v o = g τ 2 − H 2 τ 292 CHAPTER 6. SYSTEMS OF PARTICLES (b) For the given values of H = 3 km = 3000 m and τ = 90 sec, we have v o = (9 . 807 m / sec 2 )(90 sec) 2 − 3000 m 2(90 sec) = 424 . 65 m / sec = 1529 km / hr...
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This note was uploaded on 12/16/2010 for the course AME 301 at USC.
 '06
 Shiflett

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