Analytical Mech Homework Solutions 39

# Analytical Mech Homework Solutions 39 - Now sin = cos = cos...

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Now sin cos cos 42 2 42 π φπ φ     −= −− = +         2 2 max 2 2 cos cos 4 2 v R g =+ D Again using Appendix B, 22 2 cos2 cos sin 2cos 1 θθ θ = 2 max 2 21 1 cos cos 2 2 2 v R g D + = 2 2 cos 1 cos 2 v g + + D Using cos sin 2 += , () 2 max 2 1s i n i n v R g =− D 2 max i n v R g = + D 4.9 (a) Here we note that the projectile is launched “downhill” towards the target, which is located a distance h below the cannon along a line at an angle φ below the horizon. α is the angle of projection that yields maximum range, R max . We can use the results from problem 4.8 for this problem. We simply have to replace the angle φ in the above result with the angle - φ , to account for the downhill slope. Thus, we get for the downhill range … φ α R max h ( ) 2 0 2 cos sin 2 cos v R g α αϕ ϕ + = The maximum range and the angle is
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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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