%%HP: T(1)A(D)F(.);
"B5-
Calcular un cuenco 
amortiguador situado aguas
abajo de una presa de 
aliviadero para q=8m^3/s*m
=0.95;P=7.4
Ho=2m y t=3.5,=1.1

=h2-t-Z
hc=q/2g(P+Ho-hc)
hc=0
hc=0.62
=8/0.952*9.81(7.4+2-.62
=0.64***
h2=0.64/2[(1+8*1.1*8^2/9.81*0.64^3-1
h2=4.42m
h2>tsalto desplazado
Z=q^2/2g^2t^2-q^2/2g(t')^2
t'=h2=4.86
Z=1.1*8^2/2*9.81*0.95*3.5^2-
1.1*8^2/2*9.81*(1.1*4.42)^2=
0.16
d=t'-t-Z
d=4.86-3.5-0.16=1.2m
" FTL2 DROP