1. **State the problem:** 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 the fluid, and $v$ is the velocity of the fluid. 3. **Important rules:** - The density of water $\rho$ is approximately $1000\ \text{kg/m}^3$. - Velocity $v$ is given as $15\ \text{m/s}$. - The pressure calculated is the dynamic pressure exerted by the water jet on the wall. 4. **Calculate the pressure:** $$P = \frac{1}{2} \times 1000 \times (15)^2$$ $$P = 500 \times 225$$ $$P = 112500\ \text{N/m}^2$$ 5. **Interpretation:** The pressure on the wall due to the water jet is $112500\ \text{N/m}^2$ or Pascals. 6. **Note:** The problem's given answer is $2.25 \times 10^5\ \text{N/m}^2$, which is twice our calculated value. This suggests the pressure might be considered as the force per unit area from the change in momentum (impulse) of the water, doubling the dynamic pressure: Using impulse-momentum principle: $$P = \rho v^2 = 1000 \times (15)^2 = 225000\ \text{N/m}^2$$ This matches the given answer. **Final answer:** $$\boxed{2.25 \times 10^5\ \text{N/m}^2}$$