A twin-engine jet transport aircraft of gross mass $m = 50{,}000\text{ kg}$ is climbing after takeoff. At this flight point, the total available thrust from both engines is $T = 120\text{ kN}$ and total aerodynamic drag is $D = 40\text{ kN}$. Taking $g = 9.81\text{ m/s}^2$, the climb gradient (flight path angle $\gamma$ in percent, $\sin \gamma \times 100$) is closest to:

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### 1. Strategy & Theoretical Principles In a steady, unaccelerated climb along flight path angle $\gamma$: $$T - D = W \sin \gamma \implies \sin \gamma = \frac{T - D}{W}$$ The climb gradient percentage is defined as $\sin \gamma \times 100\%$ (or $\tan \gamma \times 100\% \approx \sin \gamma \times 100\%$ for small angles). ### 2. Analytical Formulation & Calculation Given: - Mass $m = 50{,}000\text{ kg} \implies W = mg = 50{,}000 \times 9.81 = 490{,}500\text{ N}$ - Thrust $T = 120\text{ kN} = 120{,}000\text{ N}$ - Drag $D = 40\text{ kN} = 40{,}000\text{ N}$ $$\sin \gamma = \frac{120{,}000 - 40{,}000}{490{,}500} = \frac{80{,}000}{490{,}500} \approx 0.163098$$ $$\text{Climb Gradient} = 0.163098 \times 100\% \approx 16.31\% \approx 16.3\%$$ ### 3. Distractor Forensics - Forgetting $g = 9.81$ and using $m = 50,000$ directly gives $80/50 = 1.6$, dimensionally mismatched. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Climb gradient is $(T - D)/W$, which is independent of flight speed. - **Standard Reference:** Anderson, J.D., *Introduction to Flight*, Chapter 6.

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