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Analytical Mech Homework Solutions 42

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

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max 1 u z h ε = ax or . This result is the correct one … Thus, ( ) ( ) max max max 2 2 2 2 .0475 3.9 u z z z ε = = 0 m 1 3.9 tan 0.821 39.4 z x α α = = = Now solve for x 0 using a relation identical to (4) … 0 0 max tan 4 h x z α = ( ) 2 0 tan x α A es gain we obtain a quadratic expr sion for 0 tan u x α = which we solve as before. This time, though, the first result for u is the correct one to use … max 0 u z h ε = and we obtain … 0 11.9 h x ft = = 0 tan α .12 The 4 x and positions of the ball vs. time are z 2 D 2 1 cos x v t θ = D 2 1 1 cos sin z v t gt θ θ = D Since x v 2 D D 1 cos 2 v θ = D D The ho i rizontal range s 2 2 1 cos sin 2 2 v R g θ θ = D D D The maximum range occurs @ 0 dR d θ = D 2 1 2cos cos v dR θ = D D 2 1 1 2 cos sin sin 2 0 2 2 2 d g θ θ θ θ θ D D D D D Thus, 2 = 2 1 1 1 cos cos2 cos sin sin 2 2 2 2 θ θ θ θ = D D D D We get θ D Using the identities:
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