Tangki Air

(A)   laju air pada lubang

\begin{equation*}
\begin{split}
\cancel {P_A} + \rho \: g \: h_A + \tfrac{1}{2} \: \rho \: v_A^2 & = \cancel {P_B} + \rho \: g \: h_B + \tfrac{1}{2} \: \rho \: v_B^2 \quad {\color {blue} P_A = P_B = P_o}\\\\
\cancel {\rho} \: g \: h_A + \tfrac{1}{2} \: \cancel {\rho} \: v_A^2 & = \cancel {\rho} \: g \: h_B + \tfrac{1}{2} \: \cancel {\rho} \: v_B^2 \quad {\color {blue} v_A \approx 0}\\\\
g \: h_A + 0 & = 0 + \tfrac 12 \:.\: v_B^2 \quad {\color {blue} \text{acuan pada titik B, maka } h_B = 0} \\\\
v_B & = \sqrt{2 \: g \: h_A} \\\\
v_B & = \sqrt{2 \:.\: 10 \:.\: 4} \\\\
v_2 & =  \sqrt{80} \\\\
v_2 & = 4 \sqrt{5} \text{ m/s}
\end{split}
\end{equation*}

(B)   nilai x

Menentukan waktu air menyentuh tanah

\begin{equation*}
\begin{split}
y & = y_o + v_{oy} \: t - \tfrac 12 \: g \: t^2 \\\\
0 & = 5 + 0 - \tfrac 12 \:.\: 10 \:.\: t^2 \\\\
t & = 1 \text{ s}
\end{split}
\end{equation*}

Menentukan jarak mendatar air di tanah

\begin{equation*}
\begin{split}
x & = v_{ox} \: t \\\\
x & = 4 \sqrt{5} \:.\: 1 \\\\
x & = 4 \sqrt{5} \text{ m}
\end{split}
\end{equation*}