Can you please write a code in python explaining Example 1. Break the problem up into a first and second reach like showed in the other 3 pictures.

Thank you!!!

Example \#1 A trapezoidal channel has the flowing characteristics: slope = 0.0016; bottom width =20ft; and side slopes =1 vertical to 2 horizontal (z=2). Water at a downstream location is an embankment where the water depth is 5.0ft just upstream of the embankment and the discharge is 400cfs. Determine the distance upstream to where the flow depth is 4.60ft. Assume =1.10. yc=2.22ft and yn=3.36ft, so subcritical flow. 5.0>yn, so M1 profile. The depth will decrease proceeding in the upstream direction because dy/dx=+ in the downstream direction. in other words, going upstream the depth will approach yn. So consider two reaches: - First reach: upstream to a depth of 5.0 to 4.8 - Second reach: upstream to a depth of 4.8 to 4.6 At y=5.0ft then A=150ft2,R=3.54ft,V= 2.67ft/s,Sf=0.00037, and E=5+ (1.1)(2.67)2/[2(32.2)]=5.123ft At y=4.8ft then A=142.1ft2,R=3.43ft,V= 2.82ft/s,Sf=0.00043, and E=4.936ft Use Sf=2.22R4/3n2v2 and Sf=21(Sf1+Sf2) Sf=21(0.00037+0.00043)=0.00040x=SoSfE2E1=0.00160.000405.1234.938=156ft At y=4.6ft then A=134.3ft2,R=3.31ft,V= 2.98ft/s,Sf=0.00051, and E=4.752ft Sf=21(0.00043+0.00051)=0.00047 and x=163ft So the distance upstream to a depth of 4.60ft is 156+163=319 ft. The procedure can be continued upstream to where the normal depth occurs