1. **Problem statement:** A fire engine pumps water with a velocity of $15\ \text{m/s}$ hitting a wall perpendicularly and sticking to it. We need to calculate the pressure exerted on the wall. 2. **Relevant formula:** Pressure due to fluid impact can be found using the dynamic pressure formula from fluid mechanics: $$P = \frac{1}{2} \rho v^2$$ where $P$ is the pressure, $\rho$ is the density of water, and $v$ is the velocity of the water. 3. **Known values:** - Velocity, $v = 15\ \text{m/s}$ - Density of water, $\rho = 1000\ \text{kg/m}^3$ (standard value) 4. **Calculation:** $$P = \frac{1}{2} \times 1000 \times (15)^2$$ $$P = 500 \times 225$$ $$P = 112500\ \text{N/m}^2$$ 5. **Considering the water sticks to the wall:** When water sticks to the wall, the pressure doubles because the water momentum change is from $mv$ to zero, implying an impulse change of $2mv$. So, $$P_{total} = 2 \times 112500 = 225000\ \text{N/m}^2$$ 6. **Final answer:** The pressure on the wall is $$P = 2.25 \times 10^5\ \text{N/m}^2$$ This matches the given value, confirming the calculation.