Let's solve the problem in **three steps**, with explanations after each step: --- ### **Step 1: Find the Helicopter's Acceleration and Speed at \( h = 110\,m \)** **Given:** - Starts from rest at A - Speed at B (\( s_{AB} = 225\,m \)) is \( v_B = 79\,m/s \) - Motion is along a straight line with constant acceleration **First, find the acceleration \( a \):** We use the kinematic equation: \[ v^2 = u^2 + 2as \] Where: - \( u = 0 \) (starts from rest) - \( v = 79\,m/s \) - \( s = 225\,m \) So, \[ (79)^2 = 0 + 2a(225) \] \[ 6241 = 450a \] \[ a = \frac{6241}{450} \approx 13.87\, m/s^2 \] **Now, find the speed at \( h = 110\,m \):** Use the same equation: \[ v^2 = 0 + 2a(110) \] \[ v^2 = 2 \times 13.87 \times 110 \] \[ v^2 = 3049.4 \] \[ v = \sqrt{3049.4} \approx 55.24\, m/s \] **Explanation:** We found the acceleration and the speed of the helicopter at the instant when \( h = 110\,m \). --- ### **Step 2: Find the Position Coordinates and R** From the diagram, when \( h = 110\,m \): - The horizontal distance from \( O \) to \( A \) is \( 490\,m \) - **Let \( x \) be the distance along the path from \( A \) to the current position \( C \)**. - At \( h = 110\,m \), \( x = h = 110\,m \) (since it’s measured along the path). Coordinates of \( C \): - \( x_C = 490 + 110 = 600\,m \) (horizontal from \( O \)) - \( y_C = 0 \) (since path is along \( x \)-axis direction) - \( z_C = \) vertical direction = \( 0 \) (since the path is in \( x-y \) plane and \( h \) is the distance along the path, not height above ground; **if h is height, clarify**) But from the diagram, it looks like \( h \) is the distance from \( A \) along the path, not a vertical height. So, from \( O \) to \( C \): - \( x_{OC} = 490 + 110 \) - \( y_{OC} = 0 \) - \( z_{OC} = 0 \) Now, let's calculate the distance \( R \) from \( O \) to \( C \): \[ R = \sqrt{(490 + 110)^2 + 0^2} = \sqrt{600^2} = 600\,m \] **Explanation:** Found the position of the helicopter and the distance from the radar to the helicopter. --- ### **Step 3: Find the Rates \( \dot{R}, \dot{\theta}, \dot{\phi} \)** - \( \dot{R} \) is the rate of change of \( R \) (radial velocity component) - \( \dot{\theta} \) is the rate of change of \( \theta \) (angle in the horizontal plane) - \( \dot{\phi} \) is the rate of change of \( \phi \) (angle in the vertical plane) From the path, the velocity \( v \) is tangent to the path (i.e., along the path from A to B). Given the geometry, the velocity vector at this instant is along the path, so: - **Radial velocity** \( \dot{R} \) is the component of \( v \) along \( R \) - **Transverse velocity** \( R\dot{\theta} \) is the component of \( v \) perpendicular to \( R \) in the horizontal plane - **Vertical velocity** \( R\dot{\phi} \) is the component of \( v \) perpendicular to \( R \) in the vertical plane But since the path is a straight line from A to B and the radar is at O, the radial component is: \[ \dot{R} = v \cos \alpha \] where \( \alpha \) is the angle between the velocity vector and the line OC. - \( \dot{\theta} = \frac{v \sin \alpha}{R} \) (for the azimuth angle) - \( \dot{\phi} = 0 \) (since the motion is in a plane, unless there is vertical motion) But given the problem is in 3D, let's consider the actual geometry for the angles: Let’s use the law of cosines to find \( \alpha \): - At \( h = 110\,m \), position C is \( 110\,m \) up the path from A, so coordinates are: - \( x_C = 490 + 110\cos \theta \) - \( y_C = 110\sin \theta \) But the path from A to B is along the y-direction, so at \( h = 110\,m \) from A: - \( x_C = 490 \) - \( y_C = 110 \) - \( z_C = 0 \) So, \[ R = \sqrt{490^2 + 110^2} = \sqrt{240100 + 12100} = \sqrt{252200} \approx 502.2\, m \] The velocity vector is along the path, which is along \( y \). - \( v_x = 0 \) - \( v_y = 55.24\, m/s \) - \( v_z = 0 \) The angle between the velocity and the vector from O to C, \( \alpha \), can be found by the dot product: \[ \cos \alpha = \frac{\vec{v} \cdot \vec{R}}{|\vec{v}||\vec{R}|} \] \[ \vec{R} = (490, 110, 0) \] \[ \vec{v} = (0, 55.24, 0) \] \[ \vec{v} \cdot \vec{R} = 0 \times 490 + 55.24 \times 110 = 6076.4 \] \[ |\vec{v}| = 55.24 \] \[ |\vec{R}| = 502.2 \] \[ \cos \alpha = \frac{6076.4}{55.24 \times 502.2} = \frac{6076.4}{27729.5} = 0.2192 \] \[ \alpha = \arccos(0.2192) \approx 77.3^\circ \] So, \[ \dot{R} = v \cos \alpha = 55.24 \times 0.2192 \approx 12.11\, m/s \] \[ R\dot{\theta} = v \sin \alpha = 55.24 \times \sin(77.3^\circ) \approx 55.24 \times 0.9757 = 53.89\, m/s \] \[ \dot{\theta} = \frac{53.89}{502.2} \approx 0.1073\, rad/s \] \[ \dot{\phi} = 0 \quad \text{(since no vertical motion)} \] --- ## **Final Answers** **At the instant when \( h = 110\,m \):** \[ \boxed{ \begin{align*} \dot{R} &\approx 12.11\, m/s \\ \dot{\theta} &\approx 0.107\, rad/s \\ \dot{\phi} &= 0 \end{align*} } \] --- ### **Summary of Steps:** 1. **Calculated the acceleration and speed at \( h = 110\,m \) using kinematics.** 2. **Found the position and distance from radar to helicopter using geometry.** 3. **Used vector analysis to find the components of velocity relative to the radar, and calculated \( \dot{R}, \dot{\theta}, \dot{\phi} \).**

Let's verify and clarify the solution step-by-step to ensure accuracy. --- ### **Step 1: Find the acceleration \(a\) and speed at \( h = 110\,m \)** **Given:** - Starting from rest at point A. - Distance from A to B: \( s_{AB} = 225\,m \). - Speed at B: \( v_B = 79\,m/s \). **Using the kinematic equation:** \[ v^2 = u^2 + 2as \] - At B: \[ 79^2 = 0 + 2a \times 225 \] \[ 6241 = 450a \] \[ a = \frac{6241}{450} \approx 13.87\,m/s^2 \] - At \( h = 110\,m \) (distance from A along the path): \[ v_{h}^2 = 0 + 2a \times 110 \] \[ v_{h}^2 = 2 \times 13.87 \times 110 \approx 3047.4 \] \[ v_{h} \approx \sqrt{3047.4} \approx 55.23\,m/s \] **_Explanation:_** We used the known acceleration to find the speed at the position corresponding to \( h=110\,m \). This provides the instantaneous velocity at that point. --- ### **Step 2: Determine position coordinates and the distance \( R \) from O to C** From the diagram: - The radar at O is at \((0,0,0)\). - The helicopter moves along the path from A to B. Assuming the path is along a straight line from A at \((490\,m, 0, 0)\) to B at \((490\,m, 225\,m, 0)\), the distance along the path is measured from A. - When \( h=110\,m \), the corresponding horizontal position along the path: \[ x = 490\,m \quad (\text{constant in this assumption}) \] \[ y = \text{distance along the path} = 110\,m \] - Height \(z\) remains zero if the path is in the horizontal plane. **Coordinates of the helicopter at this instant:** \[ C = (490, 110, 0) \] **Distance \( R \) from radar at O:** \[ R = \sqrt{490^2 + 110^2} = \sqrt{240100 + 12100} = \sqrt{252200} \approx 502.2\,m \] **_Explanation:_** The position of the helicopter at \( h=110\,m \) yields the distance \( R \) from radar O. --- ### **Step 3: Find the radial and angular velocity components** Given: - Velocity magnitude: \( v \approx 55.23\,m/s \). - Position vector from O: \( \vec{r} = (490, 110, 0) \). - The velocity vector is tangent to the path. Assuming the path is straight and along the y-axis (from A to B), the velocity vector points along the y-direction: \[ \vec{v} = (0, v, 0) = (0, 55.23, 0) \] **Calculating the rates:** - **Radial velocity \( \dot{R} \):** \[ \dot{R} = \frac{\vec{r} \cdot \vec{v}}{R} = \frac{490 \times 0 + 110 \times 55.23}{502.2} \approx \frac{6075.3}{502.2} \approx 12.11\,m/s \] - **Angular rate \( \dot{\theta} \) in the horizontal plane:** \[ \dot{\theta} = \frac{v_\perp}{R} \] where \( v_\perp \) is the component perpendicular to \( R \). Since \( v_x=0 \), \( v_y=55.23 \): \[ v_\perp = v \times \frac{\text{component perpendicular to } R}{v} \] But more straightforwardly, the horizontal component of velocity tangential to the circle centered at O is: \[ R \dot{\theta} = v \times \frac{\text{projection of } \vec{v} \text{ perpendicular to } \vec{r}}{\text{magnitude of } \vec{r}} \] Alternatively, using the vector approach: \[ \dot{\theta} = \frac{ \text{component of } \vec{v} \text{ perpendicular to } \vec{r} }{ R } \] Given the path is along y, and position vector is \((490, 110, 0)\), the component of \( \vec{v} \) perpendicular to \( \vec{r} \) is: \[ \vec{v}_\perp = \vec{v} - (\vec{v} \cdot \hat{r}) \hat{r} \] But since \( \vec{v} \) is along y: \[ \hat{r} = \frac{1}{R} (490, 110, 0) \] \[ \vec{v} \cdot \hat{r} = \frac{1}{R} (0 \times 490 + 55.23 \times 110) = \frac{6075.3}{502.2} \approx 12.11\,m/s \] Therefore: \[ \vec{v}_\perp = \vec{v} - 12.11 \times \hat{r} = (0, 55.23, 0) - 12.11 \times \left( \frac{490}{502.2}, \frac{110}{502.2}, 0 \right) \] Calculate the components: \[ 12.11 \times \frac{490}{502.2} \approx 12.11 \times 0.9753 \approx 11.8 \] \[ 12.11 \times \frac{110}{502.2} \approx 12.11 \times 0.2191 \approx 2.65 \] Thus: \[ \vec{v}_\perp \approx (0 - 11.8, 55.23 - 2.65, 0 - 0) = (-11.8, 52.58, 0) \] Magnitude: \[ |\vec{v}_\perp| = \sqrt{(-11.8)^2 + 52.58^2} \approx \sqrt{139.2 + 2765.4} \approx \sqrt{2904.6} \approx 53.9\,m/s \] Finally: \[ \dot{\theta} = \frac{|\vec{v}_\perp|}{R} \approx \frac{53.9}{502.2} \approx 0.107\, rad/s \] - **Vertical rate \( \dot{\phi} \):** Since the motion is in the horizontal plane, \( \dot{\phi} \approx 0 \). --- ### **Final summarized answers:** \[ \boxed{ \begin{aligned} & \textbf{Radial velocity } \dot{R} \approx 12.11\,m/s \\ & \textbf{Azimuth angular velocity } \dot{\theta} \approx 0.107\,rad/s \\ & \textbf{Vertical angular velocity } \dot{\phi} = 0 \end{aligned} } \] --- ### **Conclusion:** The initial solution is correct within the assumptions about the path and coordinate system. The computed values for \( \dot{R} \), \( \dot{\theta} \), and \( \dot{\phi} \) are consistent with the geometry and given data. **Final answer:** **\(\boxed{ \begin{aligned} \text{R} &\approx 502\,m \\ \dot{R} &\approx 12.11\,m/s \\ \dot{\theta} &\approx 0.107\,rad/s \\ \dot{\phi} &= 0 \end{aligned} }\)**