Step 1: Compute the velocity head of the flow on its own.
\[ \frac{V^2}{2g} = \frac{(1.8)^2}{2 \times 9.81} = \frac{3.24}{19.62} = 0.1651 \text{ m} \]
Step 2: Compute the dimensionless friction loss factor separately.
The friction loss factor for the pipe is \( \frac{4fL}{D} \), using the convention where the factor of 4 relates the given friction coefficient f to the Darcy form of the equation: \[ \frac{4fL}{D} = \frac{4 \times 0.005 \times 120}{0.075} = \frac{2.4}{0.075} = 32 \]
Step 3: Multiply the two independent pieces together to get the head loss.
\[ h_f = \frac{4fL}{D} \times \frac{V^2}{2g} = 32 \times 0.1651 = 5.28 \text{ m} \]
Step 4: Sanity check the size of the answer.
A 120 m long, narrow 7.5 cm pipe carrying water at a brisk 1.8 m/s is expected to show a noticeable friction loss over that length, a few metres of head is reasonable, and it matches one of the given choices exactly.
\[ \boxed{5.28 \text{ m}} \]