Let's break down the solution into 3 steps, with explanations after each step: --- ### **Step 1: Find the Helicopter's **Acceleration \( a \) and Speed \( v_C \) at \( h = 110\, \text{m} \)** #### **Explanation:** - The helicopter starts from rest at point \( A \), travels with constant acceleration, and has a speed of \( 79\, \text{m/s} \) when it reaches \( B \). - Distance \( AB \): \( l = \sqrt{490^2 + 225^2} = \sqrt{240100 + 50625} = \sqrt{290725} \approx 539.3\, \text{m} \) - Using \( v^2 = u^2 + 2a s \): \[ 79^2 = 0 + 2a(539.3) \] \[ a = \frac{79^2}{2 \times 539.3} = \frac{6241}{1078.6} \approx 5.787 \, \text{m/s}^2 \] Now, we need the speed at \( h = 110\, \text{m} \). Find distance from \( A \) to \( C \) (\( AC \)): - \( AC = \sqrt{490^2 + 110^2} = \sqrt{240100 + 12100} = \sqrt{252200} \approx 502.2\, \text{m} \) Find speed at \( C \): \[ v_C = \sqrt{0 + 2a(502.2)} = \sqrt{2 \times 5.787 \times 502.2} \approx \sqrt{5810.2} \approx 76.2\, \text{m/s} \] --- ### **Step 2: Find \( \dot{R} \), \( \dot{\theta} \), and \( \dot{\phi} \) using Geometry and Velocity Components** #### **Explanation:** - Let the position of \( C \) be \( (x_C, y_C, z_C) \). - At \( h = 110\, \text{m} \), \( x_C = 490\, \text{m} - AC = 490 - 502.2 \) (but actually, the path is a straight line, so use ratios). \[ \text{Let } l = \sqrt{490^2 + 225^2} \approx 539.3\, \text{m} \] \[ \frac{h}{l} = \frac{110}{539.3} \approx 0.204 \] \[ x_C = 490 - (490 \times 0.204) = 490 - 100 = 390\, \text{m} \] (But this is not quite right. Use parameterization.) The path from \( A(490, 0, 0) \) to \( B(0, 225, 0) \), so at \( h = 110\, \text{m} \): \[ x_C = 490 - \left(\frac{490}{225}\right) h = 490 - \left(\frac{490}{225}\right) \times 110 = 490 - 239.6 = 250.4\, \text{m} \] \[ y_C = 0 + h = 110\, \text{m} \] \[ z_C = 0 \] Find \( R \): \[ R = \sqrt{(250.4)^2 + (110)^2 + (580)^2} \approx \sqrt{62701.6 + 12100 + 336400} \approx \sqrt{410201.6} \approx 640.4\, \text{m} \] The velocity vector (\( \vec{v} \)) is tangent to the path from \( A \) to \( B \): \[ \vec{v} = v_C \cdot \frac{\vec{AB}}{|\vec{AB}|} \] Where \( \vec{AB} = (-490, 225, 0) \), \( |\vec{AB}| = 539.3 \). So, \[ v_x = 76.2 \cdot \frac{-490}{539.3} = -69.2\, \text{m/s} \] \[ v_y = 76.2 \cdot \frac{225}{539.3} = 31.8\, \text{m/s} \] --- ### **Step 3: Compute \( \dot{R} \), \( \dot{\theta} \), and \( \dot{\phi} \) as seen from \( O \)** #### **Explanation:** - \( \dot{R} \) is the rate of change of distance from \( O \) to \( C \). - The position vector of \( C \) from \( O \) is \( \vec{OC} = (x_C, y_C, 580) \). - \( \dot{R} = \frac{\vec{OC} \cdot \vec{v}}{R} \) \[ \vec{OC} \cdot \vec{v} = (250.4)(-69.2) + (110)(31.8) + (580)(0) = -17328.8 + 3498 = -13830.8 \] \[ \dot{R} = \frac{-13830.8}{640.4} \approx -21.6\, \text{m/s} \] - \( \dot{\theta} \) is the angular speed in the \( xy \)-plane: \[ \dot{\theta} = \frac{x_C v_y - y_C v_x}{x_C^2 + y_C^2} \] \[ \dot{\theta} = \frac{250.4 \times 31.8 - 110 \times (-69.2)}{(250.4)^2 + (110)^2} \] \[ = \frac{7962.7 + 7612}{62701.6 + 12100} \] \[ = \frac{15574.7}{74801.6} \approx 0.208\, \text{rad/s} \] - \( \dot{\phi} \) is the rate of change of elevation angle: \[ \dot{\phi} = \frac{(v_z R - \dot{R} z_C)}{R^2 - z_C^2} \] Here \( v_z = 0 \), \( z_C = 580 \), \( R = 640.4 \), \( \dot{R} = -21.6 \): \[ \dot{\phi} = \frac{0 \times 640.4 - (-21.6) \times 580}{(640.4)^2 - (580)^2} \] \[ = \frac{12528}{410112 - 336400} = \frac{12528}{73712} \approx 0.170\, \text{rad/s} \] --- ## **Final Answers** \[ \boxed{ \begin{align*} \dot{R} &\approx -21.6\, \text{m/s} \\ \dot{\theta} &\approx 0.208\, \text{rad/s} \\ \dot{\phi} &\approx 0.170\, \text{rad/s} \end{align*} } \] Let me know if you need further clarifications or a more detailed breakdown!