Use the following model for projectile motion to write the vector function that describes the path of the bomb at an altitude 40,500 feet, initial velocity of \( 660 \mathrm{~m} / \mathrm{h} \), and an angle of elevation \( \theta=0 \). \[ \begin{array}{l} r(t)=\left(v_{0} \cos \theta\right) d i+\left[h+\left(v_{0} \sin \theta\right) t-\frac{1}{2} g t^{2}\right] i \\ v_{0}=\text { Inibal velocity } \\ h=\text { initial helght } \\ 0=\text { angle of elevation } \\ g=\text { acceleration due to gravity }=32 \mathrm{ft} / \mathrm{sec} \end{array} \] Substitute the alven values accordingly and simpllify. Note: Remember to convert units of \( m / \mathrm{hr} \) to \( \mathrm{f} / \mathrm{sec} \).

To find the vector function that describes the path of the bomb, we start by converting the initial velocity from meters per hour to feet per second. Given: - Initial velocity \( v_0 = 660 \, \text{m/h} \) - Conversion factor: \( 1 \, \text{m} = 3.28084 \, \text{feet} \) - Since there are 3600 seconds in an hour: \[ v_0 = 660 \, \text{m/h} \times \frac{3.28084 \, \text{feet}}{1 \, \text{m}} \times \frac{1 \, \text{hour}}{3600 \, \text{seconds}} \approx 0.60786 \, \text{ft/s} \] Now substituting the values into the vector function: - Initial height \( h = 40,500 \, \text{ft} \) - Angle of elevation \( \theta = 0 \) degrees, which gives us: - \( \cos \theta = 1 \) - \( \sin \theta = 0 \) The formula simplifies to: \[ r(t) = \left(v_0 \cos \theta\right) t i + \left[h + \left(v_0 \sin \theta\right) t - \frac{1}{2} g t^{2}\right] j \] Substituting the values: \[ r(t) = \left(0.60786 \cdot 1\right) t i + \left[40,500 + \left(0.60786 \cdot 0\right)t - \frac{1}{2} \cdot 32 \cdot t^{2}\right] j \] \[ r(t) = 0.60786 t i + \left[40,500 - 16 t^{2}\right] j \] Thus, the vector function that describes the path of the bomb is: \[ r(t) \approx (0.60786 t) \, \mathbf{i} + (40,500 - 16 t^{2}) \, \mathbf{j} \]