The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at $v_{i, 2}=16.3 \mathrm{~m} / \mathrm{s}$ when the eagle swoops down, grabs the pigeon, and flies off. At the instant right before the attack, the eagle is flying toward the pigeon at an angle $\theta=63.9^{\circ}$ below the horizontal and a speed of $v_{i, 1}=36.5 \mathrm{~m} / \mathrm{s}$.

What is the speed $v_{\mathrm{f}}$ of the eagle immediately after it catches its prey?
\[
v_{\mathrm{f}}=
\]
\[
\mathrm{m} / \mathrm{s}
\]
What is the magnitude $\phi$ of the angle, measured from horizontal, at which the eagle is flying immediately after the strike?
\[
\phi=
\]

Step 1 ：Given the initial velocity of the eagle \(v_{i, 1} = 36.5 \, \mathrm{m/s}\), the angle at which the eagle is flying \(\theta = 63.9^\circ\), and the initial velocity of the pigeon \(v_{i, 2} = 16.3 \, \mathrm{m/s}\).

Step 2 ：Convert the angle from degrees to radians: \(\theta_{\mathrm{rad}} = \frac{\pi}{180} \times \theta = 1.1152653920243765\).

Step 3 ：Calculate the initial momentum of the eagle in the x and y directions: \(p_{i, 1_x} = 2 \times v_{i, 1} \times \cos(\theta_{\mathrm{rad}}) = 32.11555939948181\) and \(p_{i, 1_y} = 2 \times v_{i, 1} \times \sin(\theta_{\mathrm{rad}}) = 65.55601303052494\).

Step 4 ：The initial momentum of the pigeon in the y direction is \(p_{i, 2_y} = v_{i, 2} = 16.3\).

Step 5 ：Calculate the final momentum in the x and y directions: \(p_{f_x} = p_{i, 1_x} = 32.11555939948181\) and \(p_{f_y} = p_{i, 1_y} + p_{i, 2_y} = 81.85601303052493\).

Step 6 ：Calculate the final velocity in the x and y directions: \(v_{f_x} = \frac{p_{f_x}}{3} = 10.705186466493936\) and \(v_{f_y} = \frac{p_{f_y}}{3} = 27.285337676841646\).

Step 7 ：Calculate the magnitude of the final velocity: \(v_{f} = \sqrt{v_{f_x}^2 + v_{f_y}^2} = 29.310248539063586\).

Step 8 ：Calculate the direction of the final velocity: \(\phi_{\mathrm{rad}} = \arctan\left(\frac{v_{f_y}}{v_{f_x}}\right) = 1.1969089766410377\) and convert it to degrees: \(\phi = \frac{180}{\pi} \times \phi_{\mathrm{rad}} = 68.57783282285389\).

Step 9 ：Final Answer: The speed \(v_{f}\) of the eagle immediately after it catches its prey is approximately \(\boxed{29.31} \, \mathrm{m/s}\) and the magnitude \(\phi\) of the angle, measured from horizontal, at which the eagle is flying immediately after the strike is approximately \(\boxed{68.58}^\circ\).