[Q1] Maximum Height

We analyze the vertical motion of a water particle under gravity. The initial upward velocity is $v_0$, and at the maximum height $h_{max}$, the final velocity $v$ is zero. We use the kinematic equation:

$$v^2 = v_0^2 + 2as$$

Substituting $v = 0$, $a = -g$, and $s = h_{max}$:

$$0 = v_0^2 - 2gh_{max}$$

Solving for the maximum height $h_{max}$:

$$h_{max} = \frac{v_0^2}{2g}$$

Substituting the given values:

$$h_{max} = \frac{(26 \text{ m/s})^2}{2(9.8 \text{ m/s}^2)} = \frac{676}{19.6} \text{ m} \approx 34.49 \text{ m}$$

[Q2] Water Volume in the Air

In a steady state, the total volume of water in the air, $V_{air}$, is equal to the volume ejected by the hose during the time of flight, $T$, of a single water particle. The time of flight is the total time a particle takes to go up to $h_{max}$ and return to the initial height.

First, we find the time to reach the peak, $t_{up}$, using $v = v_0 + at$:

$$0 = v_0 - gt_{up}$$ $$t_{up} = \frac{v_0}{g}$$

Due to symmetry (neglecting air resistance), the time to fall back down is the same, $t_{down} = t_{up}$. The total time of flight is:

$$T = t_{up} + t_{down} = 2t_{up} = \frac{2v_0}{g}$$

The total volume of water in the air is the product of the volume flow rate $Q$ and the total time of flight $T$:

$$V_{air} = Q \times T = Q \left(\frac{2v_0}{g}\right)$$

We must use consistent units. The flow rate is $Q = 280$ L/min. We convert this to L/s:

$$Q = 280 \frac{\text{L}}{\text{min}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{280}{60} \frac{\text{L}}{\text{s}}$$

Now, we substitute the values into the derived expression for $V_{air}$:

$$V_{air} = \left(\frac{280}{60} \frac{\text{L}}{\text{s}}\right) \left(\frac{2(26 \text{ m/s})}{9.8 \text{ m/s}^2}\right) = \left(\frac{280}{60}\right) \left(\frac{52}{9.8}\right) \text{ L} \approx 24.76 \text{ L}$$