3. The plane continued to climb in altitude for 19 seconds at an angle of \( 2^{\circ} \), reaching a maximum height of 3060 feet. How much further was the plane able to travel during this time frame? (Hint: 1 mile \( =5,280 \) feet)

To determine how much further the plane traveled horizontally during its climb, we'll break down the problem step by step. ### Given: - **Climb Angle (\( \theta \))**: \( 2^{\circ} \) - **Climb Time (\( t \))**: 19 seconds - **Maximum Height Reached (\( h \))**: 3,060 feet - **Conversion**: 1 mile = 5,280 feet ### Objective: Find the horizontal distance (\( D \)) the plane traveled during the climb. ### Step 1: Calculate the Vertical Speed The vertical speed (\( V_{\text{vertical}} \)) can be calculated using the formula: \[ V_{\text{vertical}} = \frac{h}{t} = \frac{3,060 \text{ feet}}{19 \text{ seconds}} \approx 161.05 \text{ ft/s} \] ### Step 2: Relate Vertical Speed to Total Speed The vertical component of the plane's velocity is related to its total velocity (\( V \)) by the sine of the climb angle: \[ V_{\text{vertical}} = V \cdot \sin(\theta) \] \[ V = \frac{V_{\text{vertical}}}{\sin(\theta)} = \frac{161.05}{\sin(2^{\circ})} \] Calculate \( \sin(2^{\circ}) \): \[ \sin(2^{\circ}) \approx 0.0349207695 \] \[ V \approx \frac{161.05}{0.0349207695} \approx 4,612.2 \text{ ft/s} \] ### Step 3: Calculate Horizontal Speed The horizontal component of the velocity (\( V_{\text{horizontal}} \)) is: \[ V_{\text{horizontal}} = V \cdot \cos(\theta) = 4,612.2 \cdot \cos(2^{\circ}) \approx 4,612.2 \cdot 0.999390827 = 4,608.1 \text{ ft/s} \] ### Step 4: Determine Horizontal Distance Traveled \[ D = V_{\text{horizontal}} \cdot t = 4,608.1 \cdot 19 \approx 87,554 \text{ feet} \] ### Step 5: Convert Feet to Miles \[ D_{\text{miles}} = \frac{87,554 \text{ feet}}{5,280 \text{ feet/mile}} \approx 16.6 \text{ miles} \] ### **Conclusion** During the 19-second climb at a \( 2^{\circ} \) angle, the plane traveled approximately **16.6 miles** horizontally.