p0610 - at the time of the explosion). m d 2 z dt 2 = −...

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6.10. CHAPTER 6, PROBLEM 10 291 6.10 Chapter 6, Problem 10 Problem: Arocke to fmass m is launched vertically and reaches a height H with speed v o when it explodes. Part A has mass 2 5 m and, at time τ after the explosion, it strikes the ground a distance H west of the rocket’s launch point. Part B has mass 3 5 m . (a) If Part B’s height above the ground is 5 6 H when Part A strikes the ground, what is the velocity v o ? Letting g denote gravitational acceleration, express your answer as a function of τ , g and H . (b) Compute the velocity v o in km/hr for H =3 km and τ =90 sec. Solution: We appeal to Newton’s Second Law to solve for this two-particle system. (a) The only external force acting is gravity. So, the vertical position of the center of mass satisfies the following differential equation and initial conditions (with
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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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p0610 - at the time of the explosion). m d 2 z dt 2 = −...

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