Analytical Mech Homework Solutions 42

Analytical Mech Homework Solutions 42 - u zmax = h1 or u 2...

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max 1 uz h ε ≈= ax or . This result is the correct one … Thus, () ( ) max max max 2 2 2 2 .0475 3.9 z z ≈− =−= 0 m 1 3.9 tan 0.821 39.4 z x αα ==∴ = Now solve for x 0 using a relation identical to (4) … 00 max tan 4 hx z α =− () 2 0 tan x A es gain we obtain a quadratic expr sion for 0 tan ux = which we solve as before. This time, though, the first result for u is the correct one to use … max 0 h =≈ and we obtain … 0 11.9 h x ft == 0 tan .12 The 4 x and positions of the ball vs. time are z 2 D 2 1 cos xv t θ = D 2 11 cos sin zv t g t θθ D Since x v 2 DD 1 cos 2 v = The ho i rizontal range s 2 2 1 cos sin 2 2 v R g = D The maximum range occurs @ 0 dR d = D 2 1 2cos cos v dR = D D 2 1 1 2 cos sin sin 2 0 22 2 dg  D D DD D Thus, 2 = 2 1 cos cos2 cos sin sin 2 2 = D D We get D Using the identities: 2 1 =+ and sin 2
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This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

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