An open tank stands on the ground. A small hole of area 1.5 cm² in its side is 2.5 m below the…

Step 1. Speed. $\sqrt{2 \times 9.8\ \text{m/s}^2 \times 2.5\ \text{m}} = 7.0\ \text{m/s}$.

Step 2. Flow rate. $1.5 \times 10^{-4}\ \text{m}^2 \times 7.0\ \text{m/s} = 1.05 \times 10^{-3}\ \text{m}^3/\text{s}$.

Step 3. Landing. It falls for $\sqrt{\dfrac{1.6\ \text{m}}{9.8\ \text{m/s}^2}} = 0.404\ \text{s}$, so it lands $7.0\ \text{m/s} \times 0.404\ \text{s} = 2.83\ \text{m}$ away.