Dynamics_Part52

Dynamics_Part52 - Problem Set 7: Problem 3. Problem: A...

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Unformatted text preview: Problem Set 7: Problem 3. Problem: A rocket of mass m is launched vertically and reaches a height H with speed vo when it explodes. Part A has mass 1 m and, at time after the explosion, it strikes the ground a distance H 3 west of the rocket's launch point. Part B has mass 2 m. If Part B's height above the ground is 1 H 3 2 when Part A strikes the ground, what is the time ? Letting g denote gravitational acceleration, express 2 your answer as a function of vo , g and the dimensionless grouping gH/vo . Solution: We appeal to Newton's Second Law to solve. 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 t = 0 at the time of the explosion). m The solution is The center of mass is defined by z(t) = 1 m 1 2 mzA (t) + mzB (t) 3 3 d2 z = -mg, dt2 z(0) = H, z(0) = vo 1 z(t) = H + vo t - gt2 2 Now, Part A strikes the ground when zA ( ) = 0, which tells us that zB ( ) = We are given zB ( ) = 1 H. Thus, 2 3 3 1 3 H = H + vo - g 2 2 2 2 4 Solving for yields = vo g vo g 2 3 3 3 3 z( ) = H + vo - g 2 2 2 2 4 4H vo - =0 g 3 g = 2 - 2 4H 3 g + Since the solution with the minus sign is negative, we reject it. Therefore, the time is = vo 1+ g 1+ 4 gH 2 3 vo ...
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