Question 16.SP.3: Demonstrate that when taking account of fluid friction in th......

\frac{v_1^2}{2 g}-h_L=\frac{v_2^2}{2 g}, \quad where \quad h_L=k \frac{v_2^2}{2 g}

Thus:

\frac{v_1^2}{2 g}-k \frac{v_2^2}{2 g}=\frac{v_2^2}{2 g} and \quad v_1^2-k v_2^2=v_2^2

from which:                v_2=\frac{v_1}{\sqrt{1+k}}

\begin{aligned}k & =0.2, & & v_2=0.91 v_1 \\\\k & =0.6, & & v_2=0.79 v_1\end{aligned}