Analytical Mech Homework Solutions 43

# Analytical Mech Homework Solutions 43 - (a The maximum...

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(a) The maximum height occurs at 0 dz dt = cos sin 2 vg T θθ or at 1 1 cos sin 2 v T g θ = DD dH = D D or 2 22 1 cos sin v H g = D maximum at fixed D 1 dH The maximum possible height occurs @ 0 d α = Using the above trigonometric identities, 2 11 2cos sin cos cos sin sin 0 2 2 v dg  =−   D DD D D D D = we get () 1 cos sin cos sin sin sin 1 cos += = D D There ar D e 3-roots: or ( ) sin 1 cos 3cos 1 0 +− = D sin 0 = D , cos 1 = − D , 1 cos 3 = D Thus, max 18.9 os H The first two roots give minimum heights; the last gives the maximum 1 1 @ c 70 32 3 m == = D D zr s 4 The trajectory of the shell is given by Eq. 4.3.11 with r replacing .13 x 2 2 2 zg rr D ± ±± co rv = ± D sin zv = ± D r where D Thus, 2 2 tan sec gr 2 2 v Since 2 2 sec 1 tan =+ D We have: 2 tan tan 0 rz vv −+ + = D rdinates. The above equation yields two possible roots:
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