Answer:
Q = 18 m^3/sec
Explanation:
In trapezoidal channel
Area [tex]A = \frac{1}{2}(a+b)h[/tex]
where a, b are parallel side and h is distance b/w them.
Also, as per question
[tex]tan\theta = \frac{1}{2}\\\Rightarrow \theta = 26.56^{\circ}[/tex]
a = 5 m , b =5+2(2/tanθ) = 13 m
Therefore, A = 0.5(5+13)2 = 18 m^2
Discharge Q = AV = 18×1 = 18 m^3/sec
V= velocity = 1 m/s (given)
