The top air Speed of U.S F4 jet fighter is \( 2370 \mathrm{~km} / \mathrm{h} \). In a mission. it was heading to \( N 30^{\circ} E \), and the wind Speed is \( 80 \mathrm{~km} / \mathrm{h} \). W \( 45^{\circ} \mathrm{s} \). a) What is the speed of jet fighter relative to the ground?

Let the positive \(x\)‐axis point east and the positive \(y\)‐axis point north. 1. The aircraft’s airspeed is given as \(2370\,\mathrm{km/h}\) in the direction \(N30^\circ E\). This means the heading is \(30^\circ\) east of north. Its components are \[ V_{ax} = 2370\sin 30^\circ,\quad V_{ay} = 2370\cos 30^\circ. \] Since \(\sin 30^\circ=\frac{1}{2}\) and \(\cos 30^\circ\approx 0.8660\), \[ V_{ax} = 2370\left(\frac{1}{2}\right)=1185\,\mathrm{km/h}, \] \[ V_{ay} \approx 2370(0.8660)=2055.4\,\mathrm{km/h}. \] 2. The wind speed is \(80\,\mathrm{km/h}\) from \(W45^\circ S\). In meteorological notation the wind is named by the direction from which it comes. Thus the wind is coming from the direction \(45^\circ\) south of west, and it blows toward the opposite direction, which is \(E45^\circ N\). Its components are: \[ V_{wx}=80\cos45^\circ,\quad V_{wy}=80\sin45^\circ. \] Since \(\cos45^\circ=\sin45^\circ\approx 0.7071\), \[ V_{wx} \approx 80(0.7071)=56.57\,\mathrm{km/h}, \] \[ V_{wy} \approx 80(0.7071)=56.57\,\mathrm{km/h}. \] 3. The ground velocity of the jet is the vector sum of the aircraft’s airspeed and the wind speed: \[ V_{gx} = V_{ax} + V_{wx} = 1185 + 56.57\approx 1241.57\,\mathrm{km/h}, \] \[ V_{gy} = V_{ay} + V_{wy} = 2055.4 + 56.57\approx 2111.97\,\mathrm{km/h}. \] 4. The magnitude of the ground speed is \[ V_{g}=\sqrt{V_{gx}^2 + V_{gy}^2} = \sqrt{(1241.57)^2+(2111.97)^2}. \] Computing the squares: \[ (1241.57)^2\approx 1541985,\quad (2111.97)^2\approx 4451530, \] \[ V_g\approx \sqrt{1541985+4451530}=\sqrt{5993515}\approx 2448\,\mathrm{km/h}. \] Thus, the speed of the jet fighter relative to the ground is approximately \(2448\,\mathrm{km/h}\).