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Analytical Mech Homework Solutions 87

Analytical Mech Homework Solutions 87 - Total distance = h...

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Total distance 2 4 2 0 2 2 1 2 n n h h h h ε ε ε = = + + + = − + Now 0 1 n n a r = = ar , 1 r < . 2 2 2 2 0 2 1 1 2 1 1 1 n n ε ε ε ε = + − + = − + = total distance 2 2 1 1 h ε ε + = For the first fall, 2 1 2 gt h = D , so 2 h g = D t For the fall from height h : 1 2 2 h h g g ε = = t Accounting for equal rise and fall times: Total time ( ) 2 0 2 2 1 2 2 1 2 n n h h g g ε ε ε = + + = − + = + Total time 2 1 1 h g ε ε + = - v 0 /2 4m v 0 m 7.7 From eqn. 7.5.5: ( ) ( ) 1 2 1 2 2 1 1 2 m m x m m x x m m ε ε + + ′ = + ± ± ± 2 ) v t ( ) ( 1 1 1 2 1 2 1 2 m m x m m x x m m ε ε + + ′ = + ± ± ± 2 v c 1 1 4 4 4 0 5 4 4 2 4 5 c v v m m v m m m v m m m  + + + 
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