Let's break down the problem step by step: ### **Step 1: Find the Height of the Building** - Each story = 12 ft - Number of stories = 30 **Total height = 30 × 12 = 360 ft** ### **Step 2: Find the Height Half-Way Up** **Half-way up = 360 / 2 = 180 ft** --- ### **Step 3: Convert Wind Speed to Consistent Units** Given wind speed at the **top** (\(y = 360\) ft) is **75 mph**. Let's convert **75 mph** to **ft/s**: \[ 1 \text{ mph} = 5280 \text{ ft} / 3600 \text{ s} = 1.467 \text{ ft/s} \] \[ 75 \text{ mph} = 75 \times 1.467 = 110 \text{ ft/s} \] --- ### **Step 4: Wind Profile Equation \(u(y) = C_1 y^{C_2}\)** - At \(y = 360\) ft, \(u = 110\) ft/s - We want to find \(u\) at \(y = 180\) ft #### **Typical Exponent \(C_2\) for Wind Profiles** - For suburban terrain, the power-law exponent \(C_2 \approx 0.2\) (can range from 0.14 for open country to 0.33 for city centers). - We'll use **\(C_2 = 0.2\)**. #### **Find \(C_1\):** \[ u(y) = C_1 y^{C_2} \] At \(y = 360\) ft, \(u = 110\) ft/s: \[ 110 = C_1 \times (360)^{0.2} \] \[ C_1 = \frac{110}{(360)^{0.2}} \] Calculate \((360)^{0.2}\): First, \(\log_{10}(360) \approx 2.556\) \[ 360^{0.2} = 10^{0.2 \times 2.556} = 10^{0.511} \approx 3.24 \] \[ C_1 = \frac{110}{3.24} \approx 33.95 \] --- ### **Step 5: Find Wind Speed at 180 ft** \[ u(180) = 33.95 \times (180)^{0.2} \] Calculate \((180)^{0.2}\): \[ \log_{10}(180) \approx 2.255 \] \[ 180^{0.2} = 10^{0.2 \times 2.255} = 10^{0.451} \approx 2.83 \] \[ u(180) = 33.95 \times 2.83 \approx 96.1\ \text{ft/s} \] --- ### **Step 6: Calculate Dynamic Pressure \(\frac{1}{2} \rho u^2\)** - "Dynamic pressure" is usually \(\frac{1}{2} \rho u^2\) (with \(\rho\) the air density). - The problem asks for \(\frac{1}{2} u^2\), so we can use that (or \(\rho = 1\)). \[ \frac{1}{2} u^2 = \frac{1}{2} \times (96.1)^2 = \frac{1}{2} \times 9237 = 4618.5\ \text{(ft}^2/\text{s}^2) \] --- ### **Step 7: Final Answer** **Final Answer:** \[ \boxed{ \frac{1}{2} u^2 \approx 4619\ \text{ft}^2/\text{s}^2 } \] **Summary of Steps:** 1. Found building height and mid-height. 2. Converted wind speed to ft/s. 3. Used power-law profile with exponent 0.2 for suburban terrain. 4. Solved for \(C_1\). 5. Found wind speed at 180 ft. 6. Calculated dynamic pressure \(\frac{1}{2} u^2\). --- **If you want the answer in SI units (m²/s²):** - 1 ft = 0.3048 m - \(u = 96.1\ \text{ft/s} = 29.3\ \text{m/s}\) - \(\frac{1}{2} u^2 = \frac{1}{2} \times (29.3)^2 = 429\ \text{m}^2/\text{s}^2\) --- ### **Boxed Final Answers** **In ft²/s²:** \[ \boxed{ \frac{1}{2} u^2 \approx 4619\ \text{ft}^2/\text{s}^2 } \] **In m²/s²:** \[ \boxed{ \frac{1}{2} u^2 \approx 429\ \text{m}^2/\text{s}^2 } \]